I have the following code. I have been told that the time complexity of this is O(n).
But I am having a hard time understanding how. To me it seems like it is
O(n) + O(n) + O(n) = O(3n)
and the space complexity is O(3n) as well because 3 vectors that get populated to the size of n.
Is my understanding correct ?
void productofself(std::vector<int> v)
{
int multipliedProduct = 0;
std::vector<int> left;
std::vector<int> right(v.size());
std::vector<int> result;
for (int i = 0; i < v.size(); i++)
{
if (i == 0)
{
multipliedProduct = 1;
}
else
{
multipliedProduct *= v[i - 1];
}
left.push_back(multipliedProduct);
}
for (int i = v.size() - 1; i >= 0; i--)
{
if (i == v.size() - 1)
{
multipliedProduct = 1;
}
else
{
multipliedProduct *= v[i + 1];
}
right[i]=multipliedProduct;
}
for (int i = 0; i < v.size(); i++)
{
result.push_back(left[i] * right[i]);
}
}
I think you are missing the point of the asymptotic notation. That code is O(3n) but it is also O(n) and the latter is what matters.
O(3n) and O(n) are equivalent by definition.
We say f(x) is O( g(x) ) if there exist two constants m and k such that f(x) < m * g(x) for all x greater than k. Basically f(x) is O(g(x)) if for sufficiently large x, f(x) is always smaller than a constant factor times g(x).
So you can see the constant factor right in that definition: m. A scaling factor like 3 comes out in the wash. There exists an m such that 3 * n < m * n, namely any number greater than 3. So the function f(n) = 3*n is O(n).
Related
I should implement this summation in C ++. I have tried with this code, but with very high numbers up to 10 ^ 12 it takes too long.
The summation is:
For any positive integer k, let d(k) denote the number of positive divisors of k (including 1 and k itself).
For example, for the number 4: 1 has 1 divisor, 2 has two divisors, 3 has two divisors, and 4 has three divisors. So the result would be 8.
This is my code:
#include <iostream>
#include <algorithm>
using namespace std;
int findDivisors(long long n)
{
int c=0;
for(int j=1;j*j<=n;j++)
{
if(n%j==0)
{
c++;
if(j!=(n/j))
{
c++;
}
}
}
return c;
}
long long compute(long long n)
{
long long sum=0;
for(int i=1; i<=n; i++)
{
sum += (findDivisors(i));
}
return sum;
}
int main()
{
int n, divisors;
freopen("input.txt", "r", stdin);
freopen("output.txt", "w", stdout);
cin >> n;
cout << compute(n);
}
I think it's not just a simple optimization problem, but maybe I should change the algorithm entirely.
Would anyone have any ideas to speed it up? Thank you.
largest_prime_is_463035818's answer shows an O(N) solution, but the OP is trying to solve this problem
with very high numbers up to 1012.
The following is an O(N1/2) algorithm, based on some observations about the sum
n/1 + n/2 + n/3 + ... + n/n
In particular, we can count the number of terms with a specific value.
Consider all the terms n/k where k > n/2. There are n/2 of those and all are equal to 1 (integer division), so that their sum is n/2.
Similar considerations hold for the other dividends, so that we can write the following function
long long count_divisors(long long n)
{
auto sum{ n };
for (auto i{ 1ll }, k_old{ n }, k{ n }; i < k ; ++i, k_old = k)
{ // ^^^^^ it goes up to sqrt(n)
k = n / (i + 1);
sum += (k_old - k) * i;
if (i == k)
break;
sum += k;
}
return sum;
}
Here it is tested against the O(N) algorithm, the only difference in the results beeing the corner cases n = 0 and n = 1.
Edit
Thanks again to largest_prime_is_463035818, who linked the Wikipedia page about the divisor summatory function, where both an O(N) and an O(sqrt(N)) algorithm are mentioned.
An implementation of the latter may look like this
auto divisor_summatory(long long n)
{
auto sum{ 0ll };
auto k{ 1ll };
for ( ; k <= n / k; ++k )
{
sum += n / k;
}
--k;
return 2 * sum - k * k;
}
They also add this statement:
Finding a closed form for this summed expression seems to be beyond the techniques available, but it is possible to give approximations. The leading behavior of the series is given by
D(x) = xlogx + x(2γ - 1) + Δ(x)
where γ is the Euler–Mascheroni constant, and the error term is Δ(x) = O(sqrt(x)).
I used your brute force approach as reference to have test cases. The ones I used are
compute(12) == 35
cpmpute(100) == 482
Don't get confused by computing factorizations. There are some tricks one can play when factorizing numbers, but you actually don't need any of that. The solution is a plain simple O(N) loop:
#include <iostream>
#include <limits>
long long compute(long long n){
long long sum = n+1;
for (long long i=2; i < n ; ++i){
sum += n/i;
}
return sum;
}
int main()
{
std::cout << compute(12) << "\n";
std::cout << compute(100) << "\n";
}
Output:
35
482
Why does this work?
The key is in Marc Glisse's comment:
As often with this kind of problem, this sum actually counts pairs x,
y where x divides y, and the sum is arranged to count first all x
corresponding to a fixed y, but nothing says you have to keep it that
way.
I could stop here, because the comment already explains it all. Though, if it didn't click yet...
The trick is to realize that it is much simpler to count divisors of all numbers up to n rather than n-times counting divisors of individual numbers and take the sum.
You don't need to care about factorizations of eg 123123123 or 52323423 to count all divisors up to 10000000000. All you need is a change of perspective. Instead of trying to factorize numbers, consider the divisors. How often does the divisor 1 appear up to n? Simple: n-times. How often does the divisor 2 appear? Still simple: n/2 times, because every second number is divisible by 2. Divisor 3? Every 3rd number is divisible by 3. I hope you can see the pattern already.
You could even reduce the loop to only loop till n/2, because bigger numbers obviously appear only once as divisor. Though I didn't bother to go further, because the biggest change is from your O(N * sqrt(N)) to O(N).
Let's start off with some math and reduce the O(n * sq(n)) factorization to O(n * log(log(n))) and for counting the sum of divisors the overall complexity is O(n * log(log(n)) + n * n^(1/3)).
For instance:
In Codeforces himanshujaju explains how we can optimize the solution of finding divisors of a number.
I am simplifying it a little bit.
Let, n as the product of three numbers p, q, and r.
so assume p * q * r = n, where p <= q <= r.
The maximum value of p = n^(1/3).
Now we can loop over all prime numbers in a range [2, n^(1/3)]
and try to reduce the time complexity of prime factorization.
We will split our number n into two numbers x and y => x * y = n.
And x contains prime factors up to n^(1/3) and y deals with higher prime factors greater than n^(1/3).
Thus gcd(x, y) = 1.
Now define F(n) as the number of prime factors of n.
From multiplicative rules, we can say that
F(x * y) = F(x) * F(y), if gcd(x, y) = 1.
For finding F(n) => F(x * y) = F(x) * F(y)
So first find F(x) then F(y) will F(n/x)
And there will 3 cases to cover for y:
1. y is a prime number: F(y) = 2.
2. y is the square of a prime number: F(y) = 3.
3. y is a product of two distinct prime numbers: F(y) = 4.
So once we are done with finding F(x) and F(y), we are also done with finding F(x * y) or F(n).
In Cp-Algorithm there is also a nice explanation of how to count the number of divisors on a number. And also in GeeksForGeeks a nice coding example of how to count the number of divisors of a number in an efficient way. One can check the articles and can generate a nice solution to this problem.
C++ implementation
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e6 + 11;
bool prime[maxn];
bool primesquare[maxn];
int table[maxn]; // for storing primes
void SieveOfEratosthenes()
{
for(int i = 2; i < maxn; i++){
prime[i] = true;
}
for(int i = 0; i < maxn; i++){
primesquare[i] = false;
}
// 1 is not a prime number
prime[1] = false;
for(int p = 2; p * p < maxn; p++){
// If prime[p] is not changed, then
// it is a prime
if(prime[p] == true){
// Update all multiples of p
for(int i = p * 2; i < maxn; i += p){
prime[i] = false;
}
}
}
int j = 0;
for(int p = 2; p < maxn; p++) {
if (prime[p]) {
// Storing primes in an array
table[j] = p;
// Update value in primesquare[p * p],
// if p is prime.
if(p < maxn / p) primesquare[p * p] = true;
j++;
}
}
}
// Function to count divisors
int countDivisors(int n)
{
// If number is 1, then it will have only 1
// as a factor. So, total factors will be 1.
if (n == 1)
return 1;
// ans will contain total number of distinct
// divisors
int ans = 1;
// Loop for counting factors of n
for(int i = 0;; i++){
// table[i] is not less than cube root n
if(table[i] * table[i] * table[i] > n)
break;
// Calculating power of table[i] in n.
int cnt = 1; // cnt is power of prime table[i] in n.
while (n % table[i] == 0){ // if table[i] is a factor of n
n = n / table[i];
cnt = cnt + 1; // incrementing power
}
// Calculating the number of divisors
// If n = a^p * b^q then total divisors of n
// are (p+1)*(q+1)
ans = ans * cnt;
}
// if table[i] is greater than cube root of n
// First case
if (prime[n])
ans = ans * 2;
// Second case
else if (primesquare[n])
ans = ans * 3;
// Third case
else if (n != 1)
ans = ans * 4;
return ans; // Total divisors
}
int main()
{
SieveOfEratosthenes();
int sum = 0;
int n = 5;
for(int i = 1; i <= n; i++){
sum += countDivisors(i);
}
cout << sum << endl;
return 0;
}
Output
n = 4 => 8
n = 5 => 10
Complexity
Time complexity: O(n * log(log(n)) + n * n^(1/3))
Space complexity: O(n)
Thanks, #largest_prime_is_463035818 for pointing out my mistake.
I am having trouble deciding between N^2 and NlogN as the Big O? Whats throwing me off is the third nested for loop from k <=j. How do I reconcile this?
int Max_Subsequence_Sum( const int A[], const int N )
{
int This_Sum = 0, Max_Sum = 0;
for (int i=0; i<N; i++)
{
for (int j=i; j<N; j++)
{
This_Sum = 0;
for (int k=i; k<=j; k++)
{
This_Sum += A[k];
}
if (This_Sum > Max_Sum)
{
Max_Sum = This_Sum;
}
}
}
return Max_Sum;
}
This can be done with estimation or analysis. Looking at the inner most loop there are j-i operations inside the second loop. To get the total number of operations one would sum to get :
(1+N)(2 N + N^2) / 6
Making the algorithm O(N^3). To estimate one can see that there are three loops which at some point have O(N) calls thus it's O(N^3).
Let us analyze the most inner loop first:
for (int k=i; k <= j; k++) {
This_Sum += A[k];
}
Here the counter k iterates from i (inclusive) to j (inclusive), this thus means that the body of the for loop is performed j-i+1 times. If we assume that fetching the k-th number from an array is done in constant time, and the arithmetic operations (incrementing k, calculating the sum of This_Sum and A[k], and comparking k with j), then this thus runs in O(j-i).
The initialization of This_Sum and the if statement is not significant:
This_Sum = 0;
// ...
if (This_Sum > Max_Sum) {
Max_Sum = This_Sum;
}
indeed, if we can compare two numbers in constant time, and set one variable to the value hold by another value in constant time, then regardless whether the condition holds or not, the number of operations is fixed.
Now we can take a look at the loop in the middle, and abstract away the most inner loop:
for (int j=i; j < N; j++) {
// constant number of oprations
// j-i+1 operations
// constant number of operations
}
Here j ranges from i to N, so that means that the total number of operations is:
N
---
\
/ j - i + 1
---
j=i
This sum is equivalent to:
N
---
\
(N-j) * (1 - i) + / j
---
j=i
This is an arithmetic sum [wiki] and it is equivalent to:
(N - i + 1) × ((1 - i) + (i+N) / 2) = (N - i + 1) × ((N-i) / 2 + 1)
or when we expand this:
i2/2 + 3×N/2 - 3×i/2 + N2/2 - N×i + 1
So that means that we can now focus on the outer loop:
for (int i=0; i<N; i++) {
// i2/2 + 3×N/2 - 3×i/2 + N2/2 - N×i + 1
}
So now we can again calculate the number of operations with:
N
---
\
/ i2/2 + 3×N/2 - 3×i/2 + N2/2 - N×i + 1
---
i=0
We can use Faulhaber's formula [wiki] here to solve this sum, and obtain:
(N+1)×(N2+5×N+6)/6
or in expanded form:
N3/6 + N2 + 11×N/6 + 1
which is thus an O(n3) algorithm.
How to finds the number of tuplets/pairs i, j, k in array such that a[i] + a[j] = 2 * a[k]. The complexity should be O(n * logn) or O(n) since n <= 10^5.
Edit 2(important): abs(a[i]) <= 10^3.
Edit:
i, j, k must all be distinct.
Here is my code, but it's too slow, it's complexity O(is n^2 logn).
#include <bits/stdc++.h>
using namespace std;
int binarna(vector<int> a, int k){
int n = a.size();
int low = 0, high = n - 1;
bool b = false;
int mid;
while(low <= high){
mid = (low + high) / 2;
if (a[mid] == k){
b = true;
break;
}
if (k < a[mid])
high = mid - 1;
else
low = mid + 1;
}
if (b)
return mid;
else
return -1;
}
int main()
{
int n;
cin >> n;
vector<int> a(n);
for (auto& i : a)
cin >> i;
sort(a.begin(), a.end());
int sol = 0;
for (int i = 0; i < n - 1; ++i){
for (int j = i + 1; j < n; ++j){
if ((a[i] + a[j]) % 2)
continue;
int k = (a[i] + a[j]) / 2;
if (binarna(a, k) != -1)
++sol;
}
}
cout << sol << '\n';
}
The complexity can't probably be better than O(N²) because in the case of elements forming a single arithmetic progression, all pairs (i, j) with j-i even have a suitable element in the middle and the count is O(N²)*.
An O(N²) solution is as follows:
sort the array increasingly;
for every i,
set k=i and for every j>i,
increment k until 2 A[k] >= A[i] + A[j]
increment the count if equality is achieved
For a given i, j and k are monotonously increasing up to N so that the total number of operations is O(N-i). This justifies the global behavior O(N²), which is optimal.
*There is a little subtlety here as you might contradict the argument by claiming: "we can identify that the array forms an arithmetic sequence in time O(N), and from this compute the count in a single go".
But if instead of a single arithmetic sequence, we have two of them of length N/2, the quadratic behavior of the count remains even if they are intertwined. And there are at least N ways to intertwine two arithmetic sequences.
If the range of elements is much smaller than their number, it is advantageous to compress the data by means of an histogram.
The triple detection algorithm simplifies a little because k is systematically (i+j)/2. Every triple now counts for Hi.Hk.Hj instead of 1. The complexity is O(M²), where M is the size of the histogram.
Let's call D - total number of distinct values in the array. If abs(a[i]) <= 10^3, then you can't have more than 2*10^3 distinct values in the array. It means that if you are a bit smart, complexity of your algorithm becomes minimum of O(D^2*log(D)) and O(N*log(N)) which is far better than O(N^2*log(N)) and if you use smart algorithm suggested by Yves, you get minimum of O(D^2*log(D)) and O(N*log(N)).
Obviously O(N*log(N)) comes from sorting and you can't avoid it but that's OK even for N = 10^5. So how to reduce N to D in the main part of the algorithm? It is not hard. What you need is to replace the array of int values with an array of tuples (value, count) (let's call it B). It is easy to get such an array by scanning the original array after it is being sorted. The size of this new array is D (instead of N). Now you apply your algorithm or Yves improved algorithm to this array but each time you find a triplet (i,j,k) such that
2*B[k].value == B[i].value + B[j].value
you increment your total counter by
totalCount += B[k].count * B[i].count * B[j].count
Why this works? Consider the original sorted array. When you find a triplet (i,j,k) such that
2*A[k].value == A[i].value + A[j].value
You actually find 3 ranges for i, j and k such that in each range values are equal and so you can pick any number from the corresponding range. And simple combinatorics suggest the formula above.
I have written a function that takes a vector of ints, and returns the minimum difference between two elements (ie, the difference of the elements that are nearest to one another).
I am trying to work out the algorithmic complexity. The first for loop iterates over all elements of the input (I'll call this N), which is linear complexity: O(n). The inner for loop iterates over N - 1 the first time, then N - 2, N - 3, ... N - N + 2, N - N + 1, 0. So I don't think this algorithm is O(n^2). My guess is it is O(N * log(N)), is that correct?
#include <cstdlib>
#include <vector>
long int function(const std::vector<int> &Input)
{
long int MinimumDifference = -1;
for(size_t i = 0; i < Input.size(); i++)
{
for(size_t j = i + 1; j < Input.size(); j++)
{
long int Difference = abs(static_cast<long int>(Input[i] - Input[j]));
if(Difference < MinimumDifference || MinimumDifference < 0)
{
MinimumDifference = Difference;
}
}
}
return MinimumDifference;
}
Edit:
So the above is O(n^2), think I can achieve O(N * log(N)) by sorting the list first (using std::sort, which has O(N * log(N))), and then just iterate over the list to find the smallest difference.
#include <cstdlib>
#include <vector>
£include <algorithm>
void function(const std::vector<int> &Input)
{
std::vector<int> SortedInput = Input;
std::sort (SortedInput.begin(), SortedInput.end());
long int MinimumDifference = -1;
for(size_t i = 0; i < SortedInput.size() - 1; i++)
{
long int Difference = abs(static_cast<long int>(SortedInput[i] - SortedInput[i + 1]));
if(Difference < MinimumDifference || MinimumDifference < 0)
{
MinimumDifference = Difference;
}
}
return MinimumDifference;
}
A little algebraic manipulation tells us that
(n-1) + (n-2) + ... + 1 = n * (n+1) / 2
= n^2 / 2 + O(n)
so this algorithm's complexity is indeed O(n^2)
No, you're imbricating two loops of the "roughly same size", your complexity is O(n^2/2)
= O(n^2)
To convince yourself, think that sum(i) for i in [0,n] = n.(n+1)/2 = O(n^2/2) = O(n^2) which is exactly what your doing.
To do some "complexity for newbies", just count the number of imbricated loops. If they iterate on base element, multiply by n, if they reduce the problem complexity by an order (split problem in halves), multiply by log(n)
The problem is to find the n-th Catalan number mod m, where m is NOT prime, m = (10^14 + 7). Here are the list of methods that I have tried: (max N = 10,000)
Dynamic programming for table look-up, too slow
Use Catalan formula ncr(2*n, n)/(n + 1), again it wasn't fast enough due to the ncr function, can't speed up using exponentiation squaring because m is not prime.
Hardcode a table of pre-generated Catalans, but it failed due to the file size limit.
Recurrence relation C(i,k) = C(i-1,k-1) + C(i-1,k), this is way too slow
So I wonder is there any other faster algorithm to find the n-th Catalan number that I'm not aware of?
Using Dynamic Programming
void generate_catalan_numbers() {
catalan[1] = 1;
for (int i = 2; i <= MAX_NUMBERS; i++) {
for (int j = 1; j <= i - 1; j++) {
catalan[i] = (catalan[i] + ((catalan[j]) * catalan[i - j]) % MODULO) % MODULO;
}
catalan[i] = catalan[i] % MODULO;
}
}
Using original formula
ull n_choose_r(ull n, ull r) {
if (n < r)
return 0;
if (r > n/2) {
r = n - r;
}
ull result = 1;
ull common_divisor;
for (int i = 1; i <= r; ++i) {
common_divisor = gcd(result, i);
result /= common_divisor;
result *= (n - i + 1) / (i / common_divisor);
}
return result;
}
Using recurrence relation
ull n_choose_r_relation(ull n, ull r) {
for (int i = 0; i <= n + 1; ++i) {
for (int k = 0; k <= r && k <= i; ++k) {
if (k == 0 || k == i) {
ncr[i][k] = 1;
}
else {
ncr[i][k] = (ncr[i - 1][k - 1] + ncr[i - 1][k]) % MODULO;
}
}
}
return ncr[n][r];
}
From an answer I wrote just as this question about computation of nCr got closed, which ended up in the comments instead:
I'm not sure this is the absolute fastest, but it should be efficient enough. The key is that modulo multiplication decomposes, but division doesn't, so one first must reduce the fraction, as follows:
Since n <= 10000, it's very possible to build an array of size 2*n.
Use the Sieve of Eratosthenes to find and store pairs of factors for all composite numbers up to 20000). This step only needs to be done once regardless of how many Catalan numbers are to be calculated.
Make another table of size 2*n which represents the exponent of each factor.
Now, iterate the product in the Catalan formula .
Break each factor into prime factors using the sieve table, incrementing the exponent table for each term in the numerator and decrementing it for each term in the denominator.
No entry will ever end up negative.
Now, use modulo arithmetic to multiply together the non-cancelled factors.
No division operations are required at any point. Nor any fractions.
Demonstration of my method applied to multi-nCr: http://ideone.com/Weeg6
To use this for Catalan numbers, you can use this instead of the loops inside calc_combinations:
for( unsigned k = 2; k <= N; ++k ) {
factor<+1>(k+N);
factor<-1>(k);
}
The code then looks like this: http://ideone.com/ZZApk
Solution
#include <utility>
#include <vector>
std::vector< std::pair<int, int> > factor_table;
void fill_sieve( int n )
{
factor_table.resize(n+1);
for( int i = 1; i <= n; ++i )
factor_table[i] = std::pair<int, int>(i, 1);
for( int j = 2, j2 = 4; j2 <= n; (j2 += j), (j2 += ++j) ) {
if (factor_table[j].second == 1) {
int i = j;
int ij = j2;
while (ij <= n) {
factor_table[ij] = std::pair<int, int>(j, i);
++i;
ij += j;
}
}
}
}
std::vector<unsigned> powers;
template<int dir>
void factor( int num )
{
while (num != 1) {
powers[factor_table[num].first] += dir;
num = factor_table[num].second;
}
}
void calc_catalan(unsigned N)
{
powers.resize(0);
powers.resize(2*N+1);
for( unsigned k = 2; k <= N; ++k ) {
factor<+1>(k+N);
factor<-1>(k);
}
}
Test Driver
#include <iostream>
#include <cmath>
int main(void)
{
fill_sieve(20000);
unsigned N = 9913;
unsigned long long M = 1000000000007LL;
calc_catalan(N);
unsigned long long result = 1;
for( unsigned i = 0; i < powers.size(); ++i ) {
while (powers[i]--) {
result *= i;
result %= M;
}
}
std::cout << "Catalan(" << N << ") modulo " << M << " = " << result << "\n\n";
}
Completed demo: http://ideone.com/FDWfB
Here's another related question, which I answered with code and demonstration: Number of combinations (N choose R) in C++
Easy, peasy. Compute the prime factors of the binomial coefficient. A simple task using a sieve. I won't get into the rest of it, but a powermod computation is trivial, and you don't even need a divide.
For N = 10000, I get 42224403014400 in short order.
But, if you want the full number itself, the 10000'th Catalan number itself is ...
22453781249338521563359358425736057870110358621936588777329371383585
443658870053449099810271911432021020990539379958970114932732650095370271
397751300183876130693653440780258549445459994177372998459176454278220288
679699783327649549651476024591222065426709156831181207130089121989402216
517545144106669143509197596949973192167548893412063804651413496597406903
967719298471463870452875276986356795262033484770727452974197655810423629
386184662262278329466750526865120502476640878488187299740404235631962632
335108916990663560351330901464515744357084282208286669901241545533951877
777078174205283779947690623035078595904048715811899275348402286537327410
009576296851062523691528014340846065120667839872568170381150542379156626
173532955062796771718993285598391346886779480658586379448386923993317934
139425945651509102645665277040984870211604644540699508509248821099873225
565699224344151993874742555422872473424262356666363196825449089721410665
537521519676271082500130505509387186351879731113568837096419481746389018
721284533242225719341420124434480886444987373634542567071582458263380247
628252179873943804465262216365735901268165347321451279736504798992232739
106390706179212626442096326217616178171108663008963682821183764312867791
507672494716865305031842633900748973827504534625795968537648004286087039
823233370550650634239448544304798764239028734674653967478032618882557954
859328131980782727940394400855369003385513208814011609977239377877068501
893633819436630205358663340684840462204867552576509569736390978718963517
869423927523718504671005747648411794527978689778762460237949479732242725
154275831263823307362585789708343583184171797113785187466609433767144371
710845773715328364171910363978492352051901370003068055356444233141131383
192077598317531370925033378421138581148001529316546340657631162629562941
211065221871760353772365014435796695284269667873562415761642871681276498
507492541421942131281008978510862112693424595990036710403533420006771490
575482785612280198742983770649313043583275207213939274300662039637048647
395250014477941359641726047221826652916778311801541491816826072282488555
018173563867058868251361080516013361134986419403377613243853586312008767
909635869692823359899687030213634793656744420820912530014968355236934193
747181786083577435923400955703014812335311495073521773651461701750485101
119310472898683618090898735223665962918372501660743711042258315604294195
583076309209507444333462531858856911411408798540404888967120239682480627
570158137868956844950713279360385273144560292399045892610118082102910880
862332337854786916935223744892537176357434650161037841572213751901947447
479406915511862629144757855890852243043614898752155191154178797427659170
858428903659564218086017881546286273599385917718058276038925354040884258
022546721698832195059172836919416429064599278227491956109630837263590884
232587058023101145921693423507849076470763334833613166731358258440439729
023251976962577737416518794914009277934381234511794730677137605309953636
716963188964230436087118746073758080815722286112796870306754227017546055
347853334923811143440952672436342961180384459596879312187164969968096364
679341577416027452001090523659332406246454292701122715894579618818643071
139925009651888661718404932582731927646801878919152052218535889565319288
284306134970608577076704660104569794464663831193002735423564364371354521
236158069405955372080665906666149641642367693009585743888230289135078928
729184475260174446278915850624301208853693618442212023236924456444468934
014289741543223145235333811594418344798647068944904371005158995839127368
111629241573877617157577569590584624720552246920280151741755137476154967
741272080362312952750328628775530857638646138592895858764915987201920286
661490154786097488396300779244279606416541720716707237058679072236693234
932525387774462125138686406910133757255779021404876020200833761157767584
015369673586027681003369474431448843539054790848335705489738731700240579
310855452462903455809888697753847348175077261616431384533713924568807999
599683993362082982833949280082553659996487889394727840889035163412693106
865702752400579571351436509808650503057036278511515529330634352096987240
087618010503197530225589878764240330302768263496958673020211712107611762
945771002810537812467742009399047607169797035466100221770262334445478074
080845928677855301631860443068261061887109865290453732333638130446973519
286828584088203627113605849939106943614542645022903932947597417823646592
053417189520415596451505598330301782369213897762201629272201936584136036
027455748892667375417522206148332891409959866390232031014358337935412166
499617373308661369292739138448626161089231445046384163766705419698533262
040353901193260661841441922949263756492472641127072018961101915467728184
640938751407261817683231072132781927769994322689591991504965204544928105
747119997826784396172488376877215547707335474490892399544875233372674064
229287210750045834971802632275569822679385098328070604595140732389126327
092826465756212595551194678295464565601548041854366455751504169209131794
100099734293551231149329072243438440125013340293416345726479426178738686
238273833019523777019099811511419301476900607138083408535229058593795242
998150989330379630607152057165593682028276808657989133687600036850256257
973833780907105126134335912174477305526445570101413725539992976023375381
201759604514592679113676113078381084050224814280307372001545194100603017
219283437543128615425515965977881708976796492254901456997277712672653778
789696887633779923567912536882486775488103616173080561347127863398147885
811314120272830343521897029277536628882920301387371334992369039412492040
272569854478601604868543152581104741474604522753521632753090182704058850
525546680379379188800223157168606861776429258407513523623704438333489387
460217759660297923471793682082742722961582765796049294605969530190679149
426065241142453853283673009798518752237906836442958353267589634936329512
043142900668824981800672231156890228835045258196841806861681826866706774
199447245550164975361170844597908233890221446745462710788815648943858461
7793175431865532382711812960546611287516640