Problem: You are given two rectangles on a plane. The centers of both rectangles are located in the origin of coordinates (meaning the center of the rectangle's symmetry). The first rectangle's sides are parallel to the coordinate axes: the length of the side that is parallel to the Ox axis, equals w, the length of the side that is parallel to the Oy axis, equals h. The second rectangle can be obtained by rotating the first rectangle relative to the origin of coordinates by angle α.

<image>

Your task is to find the area of the region which belongs to both given rectangles. This region is shaded in the picture.

Input

The first line contains three integers w, h, α (1 ≤ w, h ≤ 106; 0 ≤ α ≤ 180). Angle α is given in degrees.

Output

In a single line print a real number — the area of the region which belongs to both given rectangles.

The answer will be considered correct if its relative or absolute error doesn't exceed 10 - 6.

Examples

Input

1 1 45


Output

0.828427125


Input

6 4 30


Output

19.668384925

Note

The second sample has been drawn on the picture above.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
double const pi = acos(-1.);
double const eps = 1e-8;
struct point {
  double x, y;
  point(double _x, double _y) { x = _x, y = _y; }
  point() {}
} p0[4], p1[4], pnt[111], res[111];
int cnt;
double w, h, a;
void rotateA(double &x, double &y) {
  double t = x * cos(a) - y * sin(a);
  y = x * sin(a) + y * cos(a);
  x = t;
}
double xmult(point p1, point p2, point p0) {
  return (p1.x - p0.x) * (p2.y - p0.y) - (p2.x - p0.x) * (p1.y - p0.y);
}
int same_side(point p1, point p2, point l1, point l2) {
  return xmult(l1, p1, l2) * xmult(l1, p2, l2) > eps;
}
int dot_online_in(point p, point l1, point l2) {
  return (((xmult(p, l1, l2)) > 0 ? (xmult(p, l1, l2)) : -(xmult(p, l1, l2))) <
          eps) &&
         (l1.x - p.x) * (l2.x - p.x) < eps && (l1.y - p.y) * (l2.y - p.y) < eps;
}
int dots_inline(point p1, point p2, point p3) {
  return (((xmult(p1, p2, p3)) > 0 ? (xmult(p1, p2, p3))
                                   : -(xmult(p1, p2, p3))) < eps);
}
int intersect_in(point u1, point u2, point v1, point v2) {
  if (!dots_inline(u1, u2, v1) || !dots_inline(u1, u2, v2))
    return !same_side(u1, u2, v1, v2) && !same_side(v1, v2, u1, u2);
  return dot_online_in(u1, v1, v2) || dot_online_in(u2, v1, v2) ||
         dot_online_in(v1, u1, u2) || dot_online_in(v2, u1, u2);
}
point intersection(point u1, point u2, point v1, point v2) {
  point ret = u1;
  double t = ((u1.x - v1.x) * (v1.y - v2.y) - (u1.y - v1.y) * (v1.x - v2.x)) /
             ((u1.x - u2.x) * (v1.y - v2.y) - (u1.y - u2.y) * (v1.x - v2.x));
  ret.x += (u2.x - u1.x) * t;
  ret.y += (u2.y - u1.y) * t;
  return ret;
}
bool mult(point sp, point ep, point op) {
  return (sp.x - op.x) * (ep.y - op.y) >= (ep.x - op.x) * (sp.y - op.y);
}
bool operator<(const point &l, const point &r) {
  return l.y < r.y || (fabs(l.y - r.y) < eps && l.x < r.x);
}
int graham(point pnt[], int n, point res[]) {
  int i, len, k = 0, top = 1;
  sort(pnt, pnt + n);
  for (i = 0; i < 3; i++) res[i] = pnt[i];
  if (n < 3) return n;
  for (i = 2; i < n; i++) {
    while (top && mult(pnt[i], res[top], res[top - 1])) top--;
    res[++top] = pnt[i];
  }
  len = top;
  res[++top] = pnt[n - 2];
  for (i = n - 3; i >= 0; i--) {
    while (top != len && mult(pnt[i], res[top], res[top - 1])) top--;
    res[++top] = pnt[i];
  }
  return top;
}
int main() {
  double x1, y1, x2, y2, x3, y3, x4, y4;
  while (~scanf("%lf %lf %lf", &w, &h, &a)) {
    if (a == 0 || a == 180) {
      printf("%.10lf\n", w * h);
      continue;
    }
    a = a * pi / 180.;
    x1 = 0.5 * w, y1 = 0.5 * h;
    p0[0] = point(x1, y1);
    rotateA(x1, y1);
    p1[0] = point(x1, y1);
    x2 = -0.5 * w, y2 = 0.5 * h;
    p0[1] = point(x2, y2);
    rotateA(x2, y2);
    p1[1] = point(x2, y2);
    x3 = -0.5 * w, y3 = -0.5 * h;
    p0[2] = point(x3, y3);
    rotateA(x3, y3);
    p1[2] = point(x3, y3);
    x4 = 0.5 * w, y4 = -0.5 * h;
    p0[3] = point(x4, y4);
    rotateA(x4, y4);
    p1[3] = point(x4, y4);
    cnt = 0;
    for (int i = 0; i < 4; ++i) {
      for (int j = 0; j < 4; ++j) {
        if (intersect_in(p0[i], p0[(i + 1) % 4], p1[j], p1[(j + 1) % 4])) {
          pnt[cnt++] =
              intersection(p0[i], p0[(i + 1) % 4], p1[j], p1[(j + 1) % 4]);
        }
      }
    }
    pnt[cnt] = pnt[0];
    int g = graham(pnt, cnt, res);
    double ret = 0;
    for (int i = 0; i < g - 1; ++i) {
      ret += xmult(res[i], res[i + 1], res[0]);
    }
    ret = fabs(ret) * 0.5;
    printf("%.10lf\n", ret);
  }
  return 0;
}