calling glm::unproject() correctly, confused - c++

I'm trying to use glm::unproject() to convert my SDL mouse coordinates into a world position vector, on the x/z-plane. Basically I want to figure out which "x/z" coordinate the user clicked on with a mouse.
From other stack overflow answers I came up needing to call glm::unproject(). I think I'm passing it the wrong arguments, because the values I'm getting back for the world position (printed std std::cerr) aren't world position values as I would expect.
Am I constructing the arguments to glm::unproject() correctly below? Specifically should I be combing the camera's world position and the view matrix (computed using glm::lookAt) to compute the modelview matrix passed into glm::unproject?
struct Dimensions {
int x, y, w, h;
};
glm::mat4
Camera::view_matrix() const
{
// VIEW matrix is created by looking at some target member
auto const& target = target_->translation;
auto const position_xyz = world_position();
glm::vec3 const UP{0, 1, 0};
return glm::lookAt(position_xyz, target, UP);
}
glm::mat4
Camera::projection_matrix() const
{
auto const fov = glm::radians(90.0f);
return glm::perspective(fov, 4.0f/3.0f, 0.1f, 200.0f);
}
glm::vec3
calculate_worldpos(Camera const& camera, int const mouse_x, int const mouse_y)
{
float const width = 1024.0f, height = 768.0f;
glm::vec4 const viewport = glm::vec4(0.0f, 0.0f, width, height);
glm::mat4 const modelview = camera.view_matrix();
glm::mat4 const projection = camera.projection_matrix();
float z = 0.0;
glm::vec3 screenPos = glm::vec3(mouse_x, height - mouse_y - 1, z);
std::cerr << "screenpos: xyz: '" << glm::to_string(screenPos) << "'\n";
glm::vec3 worldPos = glm::unProject(screenPos, modelview, projection, viewport);
std::cerr << "worldpos: xyz: '" << glm::to_string(worldPos) << "'\n";
return worldPos;
}
In the image below, I have the follow setup.
camera lookAt target = (0, 0, 0)
camera world position = (-0.009, 5.107, -0.368)
(mouse_x, mouse_y, mouse_z) = (286, 393, 0)
If you look at the image below, you can see that my mouse is hovering over the world position (3, 0, 0) as shown by the grid. I would expect calculating the world position of my mouse (as shown in the picture) would return me the vector (3, 0, 0). It does not, instead I get the vector: (0.049, 5.007, -0.360).
Does anyone see where I might be going wrong? I'm assuming I'm making some kind of incorrect assumption somewhere.

Your assumption is wrong: glm::unproject returns the worldspace coordinates of the input given by a xy-position in pixel coordinates and a z-coordinate storing the depth value. On every pixel on the screen, there is an infinite number of points in worldspace that project to this pixel (All that lie on the ray going from the projecting center through this pixel). Which one you want is identified by choosing the depth coordinate which than results in one specific point on this ray. Choosing z = 0 means that the result will always be a point on the near-plane of the camera.
What you are actually looking for is the intersection of this ray (going through the camera position and the calculated point) and the xz-plane (where y=0).
The ray is given by the two points on it (camera position C, near plane point P) as follows:
-0.009 0.058
C + l * (P-C) = ( 5.107 ) + l * ( -0.100 )
-0.368 0.008
, where l is a free variable.
As already said, we are looking for the intersection point (a,b) with the y=0 plane, thus we can formulate the following equation:
-0.009 0.058 a
( 5.107 ) + l * ( -0.100 ) = ( 0 )
-0.368 0.008 b
Solving the y-equation (5.107 + l * -0.1 = 0) for l results in l = 51.07. Pasting back in the equations for x and z yields:
a = -0.009 + 51.07 * 0.058 = 2.95306
b = -0.368 + 51.07 * 0.008 = 0.04056
Which is close to the expected worldspace position. The difference is most probably given by the fact that you just showed rounded numbers in the question. For accuracy reasons, I would also not calculate a point on the near-plane but one on the far plane (z=1) since the near-plane distance is usually quite small and could lead to numerical issues.
Conclusion: All values supplied are correct, but you were just not calculating what you expected.

Related

Get 3D model coordinate with 2D screen coordinates gluUnproject

I try to get the 3D coordinates of my OpenGL model. I found this code in the forum, but I donĀ“t understand how the collision is detected.
-(void)receivePoint:(CGPoint)loke
{
GLfloat projectionF[16];
GLfloat modelViewF[16];
GLint viewportI[4];
glGetFloatv(GL_MODELVIEW_MATRIX, modelViewF);
glGetFloatv(GL_PROJECTION_MATRIX, projectionF);
glGetIntegerv(GL_VIEWPORT, viewportI);
loke.y = (float) viewportI[3] - loke.y;
float nearPlanex, nearPlaney, nearPlanez, farPlanex, farPlaney, farPlanez;
gluUnProject(loke.x, loke.y, 0, modelViewF, projectionF, viewportI, &nearPlanex, &nearPlaney, &nearPlanez);
gluUnProject(loke.x, loke.y, 1, modelViewF, projectionF, viewportI, &farPlanex, &farPlaney, &farPlanez);
float rayx = farPlanex - nearPlanex;
float rayy = farPlaney - nearPlaney;
float rayz = farPlanez - nearPlanez;
float rayLength = sqrtf((rayx*rayx)+(rayy*rayy)+(rayz*rayz));
//normalizing rayVector
rayx /= rayLength;
rayy /= rayLength;
rayz /= rayLength;
float collisionPointx, collisionPointy, collisionPointz;
for (int i = 0; i < 50; i++)
{
collisionPointx = rayx * rayLength/i*50;
collisionPointy = rayy * rayLength/i*50;
collisionPointz = rayz * rayLength/i*50;
}
}
In my opinion there a break condition missing. When do I find the collisionPoint?
Another question is:
How do I manipulate the texture at these collision point? I think that I need the corresponding vertex!?
best regards
That code takes the ray from your near clipping place to your far at the position of your loke then partitions it in 50 and interpolates all the possible location of your point in 3D along this ray. At the exit of the loop, in the original code you posted, collisionPointx, y and z is the value of the far most point. There is no "collision" test in that code. you actually need to test your 3D coordinates against a 3D object you want to collide with.

Precision issue - viewpoint far from origin - OpenGL C++

I have a camera class for controlling the camera, with the main function:
void PNDCAMERA::renderMatrix()
{
float dttime=getElapsedSeconds();
GetCursorPos(&cmc.p_cursorPos);
ScreenToClient(hWnd, &cmc.p_cursorPos);
double d_horangle=((double)cmc.p_cursorPos.x-(double)cmc.p_origin.x)/(double)screenWidth*PI;
double d_verangle=((double)cmc.p_cursorPos.y-(double)cmc.p_origin.y)/(double)screenHeight*PI;
cmc.horizontalAngle=d_horangle+cmc.d_horangle_prev;
cmc.verticalAngle=d_verangle+cmc.d_verangle_prev;
if(cmc.verticalAngle>PI/2) cmc.verticalAngle=PI/2;
if(cmc.verticalAngle<-PI/2) cmc.verticalAngle=-PI/2;
changevAngle(cmc.verticalAngle);
changehAngle(cmc.horizontalAngle);
rightVector=glm::vec3(sin(horizontalAngle - PI/2.0f),0,cos(horizontalAngle - PI/2.0f));
directionVector=glm::vec3(cos(verticalAngle) * sin(horizontalAngle), sin(verticalAngle), cos(verticalAngle) * cos(horizontalAngle));
upVector=glm::vec3(glm::cross(rightVector,directionVector));
glm::normalize(upVector);
glm::normalize(directionVector);
glm::normalize(rightVector);
if(moveForw==true)
{
cameraPosition=cameraPosition+directionVector*(float)C_SPEED*dttime;
}
if(moveBack==true)
{
cameraPosition=cameraPosition-directionVector*(float)C_SPEED*dttime;
}
if(moveRight==true)
{
cameraPosition=cameraPosition+rightVector*(float)C_SPEED*dttime;
}
if(moveLeft==true)
{
cameraPosition=cameraPosition-rightVector*(float)C_SPEED*dttime;
}
glViewport(0,0,screenWidth,screenHeight);
glScissor(0,0,screenWidth,screenHeight);
projection_matrix=glm::perspective(60.0f, float(screenWidth) / float(screenHeight), 1.0f, 40000.0f);
view_matrix = glm::lookAt(
cameraPosition,
cameraPosition+directionVector,
upVector);
gShader->bindShader();
gShader->sendUniform4x4("model_matrix",glm::value_ptr(model_matrix));
gShader->sendUniform4x4("view_matrix",glm::value_ptr(view_matrix));
gShader->sendUniform4x4("projection_matrix",glm::value_ptr(projection_matrix));
gShader->sendUniform("camera_position",cameraPosition.x,cameraPosition.y,cameraPosition.z);
gShader->sendUniform("screen_size",(GLfloat)screenWidth,(GLfloat)screenHeight);
};
It runs smooth, I can control the angle with my mouse in X and Y directions, but not around the Z axis (the Y is the "up" in world space).
In my rendering method I render the terrain grid with one VAO call. The grid itself is a quad as the center (highes lod), and the others are L shaped grids scaled by powers of 2. It is always repositioned before the camera, scaled into world space, and displaced by a heightmap.
rcampos.x = round((camera_position.x)/(pow(2,6)*gridscale))*(pow(2,6)*gridscale);
rcampos.y = 0;
rcampos.z = round((camera_position.z)/(pow(2,6)*gridscale))*(pow(2,6)*gridscale);
vPos = vec3(uv.x,0,uv.y)*pow(2,LOD)*gridscale + rcampos;
vPos.y = texture(hmap,vPos.xz/horizontal_scale).r*vertical_scale;
The problem:
The camera starts at the origin, at (0,0,0). When I move it far away from that point, it causes the rotation along the X axis discontinuous. It feels like the mouse cursor was aligned with a grid in screen space, and only the position at grid points were recorded as the cursor movement.
I've also recorded the camera position when it gets pretty noticeable, it's about at 1,000,000 from the origin in X or Z directions. I've noticed that this 'lag' increases linearly with distance, (from the origin).
There is also a little Z-fighting at this point(or similar effect), even if I use a single plane with no displacement, and no planes can overlap. (I use tessellation shaders and render patches.) Black spots appear on the patches. May be caused by fog:
float fc = (view_matrix*vec4(Pos,1)).z/(view_matrix*vec4(Pos,1)).w;
float fResult = exp(-pow(0.00005f*fc, 2.0));
fResult = clamp(fResult, 0.0, 1.0);
gl_FragColor = vec4(mix(vec4(0.0,0.0,0.0,0),vec4(n,1),fResult));
Another strange behavior is the little rotation by the Z axis, this increases with distance too, but I don't use this kind of rotation.
Variable formats:
The vertices are unsigned short format, the indexes are in unsigned int format.
The cmc struct is the camera/cursor struct with double variables.
PI and C_SPEED are #define constants.
Additional information:
The grid is created with the above mentioned ushort array, with the spacing of 1. In the shader I scale it with a constant, then use tessellation to achieve the best performance and the largest view distance.
The final position of a vertex is calculated in the tessellation evaluation shader.
mat4 MVP = projection_matrix*view_matrix*model_matrix;
As you could see I send my matrices to the shader with the glm library.
+Q:
How could the length of a float (or any other format) cause this kind of 'precision loss', or whatever causes the problem. The view_matrix could be a cause of this, but I still cannot output it on the screen at runtime.
PS: I don't know If this helps, but the view matrix at about the 'lag start location' is
-0.49662 -0.49662 0.863129 0
0.00514956 0.994097 0.108373 0
-0.867953 0.0582648 -0.493217 0
1.62681e+006 16383.3 -290126 1
EDIT
Comparing the camera position and view matrix:
view matrix = 0.967928 0.967928 0.248814 0
-0.00387854 0.988207 0.153079 0
-0.251198 -0.149134 0.956378 0
-2.88212e+006 89517.1 -694945 1
position = 2.9657e+006, 6741.52, -46002
It's a long post so I might not answer everything.
I think it is most likely precision issue. Lets start with the camera rotation problem. I think the main problem is here
view_matrix = glm::lookAt(
cameraPosition,
cameraPosition+directionVector,
upVector);
As you said, position is quite a big number like 2.9657e+006 - and look what glm does in glm::lookAt:
GLM_FUNC_QUALIFIER detail::tmat4x4<T> lookAt
(
detail::tvec3<T> const & eye,
detail::tvec3<T> const & center,
detail::tvec3<T> const & up
)
{
detail::tvec3<T> f = normalize(center - eye);
detail::tvec3<T> u = normalize(up);
detail::tvec3<T> s = normalize(cross(f, u));
u = cross(s, f);
In your case, eye and center are these big (very similar) numbers and then glm subtracts them to compute f. This is bad, because if you subtract two almost equal floats, the most significant digits are set to zero, which leaves you with the insignificant (most erroneous) digits. And you use this for further computations, which only emphasizes the error. Check this link for some details.
The z-fighting is similar issue. Z-buffer is not linear, it has the best resolution near the camera because of the perspective divide. The z-buffer range is set according to your near and far clipping plane values. You always want to have the smallest possible ration between far and near values (generally far/near should not be greater than 30000). There is a very good explanation of this on the openGL wiki, I suggest you read it :)
Back to the camera issue - first, I would consider if you really need such a huge scene. I don't think so, but if yes, you could try computing your view matrix differently, compute rotation and translation separately, which could help your case. The way I usually handle camera:
glm::vec3 cameraPos;
glm::vec3 cameraRot;
glm::vec3 cameraPosLag;
glm::vec3 cameraRotLag;
int ox, oy;
const float inertia = 0.08f; //mouse inertia
const float rotateSpeed = 0.2f; //mouse rotate speed (sensitivity)
const float walkSpeed = 0.25f; //walking speed (wasd)
void updateCameraViewMatrix() {
//camera inertia
cameraPosLag += (cameraPos - cameraPosLag) * inertia;
cameraRotLag += (cameraRot - cameraRotLag) * inertia;
// view transform
g_CameraViewMatrix = glm::rotate(glm::mat4(1.0f), cameraRotLag[0], glm::vec3(1.0, 0.0, 0.0));
g_CameraViewMatrix = glm::rotate(g_CameraViewMatrix, cameraRotLag[1], glm::vec3(0.0, 1.0, 0.0));
g_CameraViewMatrix = glm::translate(g_CameraViewMatrix, cameraPosLag);
}
void mousePositionChanged(int x, int y) {
float dx, dy;
dx = (float) (x - ox);
dy = (float) (y - oy);
ox = x;
oy = y;
if (mouseRotationEnabled) {
cameraRot[0] += dy * rotateSpeed;
cameraRot[1] += dx * rotateSpeed;
}
}
void keyboardAction(int key, int action) {
switch (key) {
case 'S':// backwards
cameraPos[0] -= g_CameraViewMatrix[0][2] * walkSpeed;
cameraPos[1] -= g_CameraViewMatrix[1][2] * walkSpeed;
cameraPos[2] -= g_CameraViewMatrix[2][2] * walkSpeed;
break;
...
}
}
This way, the position would not affect your rotation. I should add that I adapted this code from NVIDIA CUDA samples v5.0 (Smoke Particles), I really like it :)
Hope at least some of this helps.

Compute mesh vertices from a Plane

I would like to draw my Plane with OpenGL to debug my program but I don't know how I can do that (I'm not very good in math).
I've got a Plane with 2 attributs:
A constant
A normal
Here is what i've got:
////////////////////////////////////////////////////////////
Plane::Plane( const glm::vec3& a, const glm::vec3& b, const glm::vec3& c )
{
glm::vec3 edge1 = b - a;
glm::vec3 edge2 = c - a;
this->normal = glm::cross(edge1, edge2);
this->constant = -glm::dot( this->normal, a );
this->normalize();
}
////////////////////////////////////////////////////////////
Plane::Plane( const glm::vec4& values )
{
this->normal = glm::vec3( values.x, values.y, values.z );
this->constant = values.w;
}
////////////////////////////////////////////////////////////
Plane::Plane( const glm::vec3& normal, const float constant ) :
constant (constant),
normal (normal)
{
}
////////////////////////////////////////////////////////////
Plane::Plane( const glm::vec3& normal, const glm::vec3& point )
{
this->normal = normal;
this->constant = -glm::dot(normal, point);
this->normalize();
}
I would like to draw it to see if everything is ok. How I can do that?
(I need to compute vertices and indices to draw it)
When you want to draw, you need to find two vectors that are perpendecular to normal and a point on the plane. That's not so hard. First, let's get a vector that is not normal. Call it some_vect. For example:
if normal == [0, 0, 1]
some_vect = [0, 1, 0]
else
some_vect = [0, 0, 1]
Then, calculating vect1 = cross(normal, some_vect) would give you a vector perpendecular to normal. Calculating vect2 = cross(normal, vect1) would give you another vector that is perpendecular to normal.
Having two perpendecular vectors vect1 and vect2 and one point on the plane, drawing the plane becomes trivial. For example the sqaure with the following four points (remember to normalize the vectors):
point + vect1 * SIZE
point + vect2 * SIZE
point - vect1 * SIZE
point - vect2 * SIZE
where point is a point on the plane. If your constant is distance from the origin, then one point would be constant * normal.
The difficulty with drawing a plane is that it's an infinite surface; i.e. by definition it has no edges or vertices. If you want to show the plane in a typical polygonal fashion then you'll have to crop it to a particular area, such as a square.
A fairly easy approach is this:
Pick an arbitrary unit vector which is perpendicular to the normal. Store it as v1.
Use the cross product of v1 and the plane normal to get v2.
Negate v1 to get v3.
Negate v2 to get v4.
The points v1-4 now express the four corners of a square which has the same orientation as your plane. All you need to do is multiply them up to whatever size you want, and draw it relative to any point on your plane.

OpenGL Frustum visibility test with sphere : Far plane not working

I am doing a program to test sphere-frustum intersection and being able to determine the sphere's visibility. I am extracting the frustum's clipping planes into camera space and checking for intersection. It works perfectly for all planes except the far plane and I cannot figure out why. I keep pulling the camera back but my program still claims the sphere is visible, despite it having been clipped long ago. If I go far enough it eventually determines that it is not visible, but this is some distance after it has exited the frustum.
I am using a unit sphere at the origin for the test. I am using the OpenGL Mathematics (GLM) library for vector and matrix data structures and for its built in math functions. Here is my code for the visibility function:
void visibilityTest(const struct MVP *mvp) {
static bool visLastTime = true;
bool visThisTime;
const glm::vec4 modelCenter_worldSpace = glm::vec4(0,0,0,1); //at origin
const int negRadius = -1; //unit sphere
//Get cam space model center
glm::vec4 modelCenter_cameraSpace = mvp->view * mvp->model * modelCenter_worldSpace;
//---------Get Frustum Planes--------
//extract projection matrix row vectors
//NOTE: since glm stores their mats in column-major order, we extract columns
glm::vec4 rowVec[4];
for(int i = 0; i < 4; i++) {
rowVec[i] = glm::vec4( mvp->projection[0][i], mvp->projection[1][i], mvp->projection[2][i], mvp->projection[3][i] );
}
//determine frustum clipping planes (in camera space)
glm::vec4 plane[6];
//NOTE: recall that indices start at zero. So M4 + M3 will be rowVec[3] + rowVec[2]
plane[0] = rowVec[3] + rowVec[2]; //near
plane[1] = rowVec[3] - rowVec[2]; //far
plane[2] = rowVec[3] + rowVec[0]; //left
plane[3] = rowVec[3] - rowVec[0]; //right
plane[4] = rowVec[3] + rowVec[1]; //bottom
plane[5] = rowVec[3] - rowVec[1]; //top
//extend view frustum by 1 all directions; near/far along local z, left/right among local x, bottom/top along local y
// -Ax' -By' -Cz' + D = D'
plane[0][3] -= plane[0][2]; // <x',y',z'> = <0,0,1>
plane[1][3] += plane[1][2]; // <0,0,-1>
plane[2][3] += plane[2][0]; // <-1,0,0>
plane[3][3] -= plane[3][0]; // <1,0,0>
plane[4][3] += plane[4][1]; // <0,-1,0>
plane[5][3] -= plane[5][1]; // <0,1,0>
//----------Determine Frustum-Sphere intersection--------
//if any of the dot products between model center and frustum plane is less than -r, then the object falls outside the view frustum
visThisTime = true;
for(int i = 0; i < 6; i++) {
if( glm::dot(plane[i], modelCenter_cameraSpace) < static_cast<float>(negRadius) ) {
visThisTime = false;
}
}
if(visThisTime != visLastTime) {
printf("Sphere is %s visible\n", (visThisTime) ? "" : "NOT " );
visLastTime = visThisTime;
}
}
The polygons appear to be clipped by the far plane properly so it seems that the projection matrix is set up properly, but the calculations make it seem like the plane is way far out. Perhaps I am not calculating something correctly or have a fundamental misunderstanding of the calculations that are required?
The calculations that deal specifically with the far clipping plane are:
plane[1] = rowVec[3] - rowVec[2]; //far
and
plane[1][3] += plane[1][2]; // <0,0,-1>
I'm setting the plane to be equal to the 4th row (or in this case column) of the projection matrix - the 3rd row of the projection matrix. Then I'm extending the far plane one unit further (due to the sphere's radius of one; D' = D - C(-1) )
I've looked over this code many times and I can't see why it shouldn't work. Any help is appreciated.
EDIT:
I can't answer my own question as I don't have the rep, so I will post it here.
The problem was that I wasn't normalizing the plane equations. This didn't seem to make much of a difference for any of the clip planes besides the far one, so I hadn't even considered it (but that didn't make it any less wrong). After normalization everything works properly.

Ray tracing vectors

So I decided to write a ray tracer the other day, but I got stuck because I forgot all my vector math.
I've got a point behind the screen (the eye/camera, 400,300,-1000) and then a point on the screen (a plane, from 0,0,0 to 800,600,0), which I'm getting just by using the x and y values of the current pixel I'm looking for (using SFML for rendering, so it's something like 267,409,0)
Problem is, I have no idea how to cast the ray correctly. I'm using this for testing sphere intersection(C++):
bool SphereCheck(Ray& ray, Sphere& sphere, float& t)
{ //operator * between 2 vec3s is a dot product
Vec3 dist = ray.start - sphere.pos; //both vec3s
float B = -1 * (ray.dir * dist);
float D = B*B - dist * dist + sphere.radius * sphere.radius; //radius is float
if(D < 0.0f)
return false;
float t0 = B - sqrtf(D);
float t1 = B + sqrtf(D);
bool ret = false;
if((t0 > 0.1f) && (t0 < t))
{
t = t0;
ret = true;
}
if((t1 > 0.1f) && (t1 < t))
{
t = t1;
ret = true;
}
return ret;
}
So I get that the start of the ray would be the eye position, but what is the direction?
Or, failing that, is there a better way of doing this? I've heard of some people using the ray start as (x, y, -1000) and the direction as (0,0,1) but I don't know how that would work.
On a side note, how would you do transformations? I'm assuming that to change the camera angle you just adjust the x and y of the camera (or the screen if you need a drastic change)
The parameter "ray" in the function,
bool SphereCheck(Ray& ray, Sphere& sphere, float& t)
{
...
}
should already contain the direction information and with this direction you need to check if the ray intersects the sphere or not. (The incoming "ray" parameter is the vector between the camera point and the pixel the ray is sent.)
Therefore the local "dist" variable seems obsolete.
One thing I can see is that when you create your rays you are not using the center of each pixel in the screen as the point for building the direction vector. You do not want to use just the (x, y) coordinates on the grid for building those vectors.
I've taken a look at your sample code and the calculation is indeed incorrect. This is what you want.
http://www.csee.umbc.edu/~olano/435f02/ray-sphere.html (I took this course in college, this guy knows his stuff)
Essentially it means you have this ray, which has an origin and direction. You have a sphere with a point and a radius. You use the ray equation and plug it into the sphere equation and solve for t. That t is the distance between the ray origin and the intersection point on the spheres surface. I do not think your code does this.
So I get that the start of the ray would be the eye position, but what is the direction?
You have camera defined by vectors front, up, and right (perpendicular to each other and normalized) and "position" (eye position).
You also have width and height of viewport (pixels), vertical field of view (vfov) and horizontal field of view (hfov) in degrees or radians.
There are also 2D x and y coordinates of pixel. X axis (2D) points to the right, Y axis (2D) points down.
For a flat screen ray can be calculated like this:
startVector = eyePos;
endVector = startVector
+ front
+ right * tan(hfov/2) * (((x + 0.5)/width)*2.0 - 1.0)
+ up * tan(vfov/2) * (1.0 - ((y + 0.5f)/height)*2.0);
rayStart = startVector;
rayDir = normalize(endVector - startVector);
That assumes that screen plane is flat. For extreme field of view angles (fov >= 180 degreess) you might want to make screen plane spherical, and use different formulas.
how would you do transformations
Matrices.