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;
long long MOD_EXPO(long long b, long long p, long long m) {
  if (p == 0) return 1;
  long long ret = MOD_EXPO(b, p / 2, m) % m;
  ret = (ret * ret) % m;
  return ((p & 1) ? (ret * b) % m : ret % m);
}
long long POWER(long long N, long long K) {
  long long i, ans = 1;
  for (i = 1; i <= K; i++) ans *= N;
  return ans;
}
int SET(int N, int pos) { return (N | (1 << pos)); }
int RESET(int N, int pos) { return (N & !(1 << pos)); }
bool CHECK(int N, int pos) { return (N & (1 << pos)); }
int dx4[] = {1, -1, 0, 0};
int dy4[] = {0, 0, 1, -1};
int dx6[] = {0, 0, 1, -1, 0, 0};
int dy6[] = {1, -1, 0, 0, 0, 0};
int dz6[] = {0, 0, 0, 0, 1, -1};
int dx8[] = {1, -1, 0, 0, -1, 1, -1, 1};
int dy8[] = {0, 0, 1, -1, 1, 1, -1, -1};
int dkx8[] = {-1, 1, -1, 1, -2, -2, 2, 2};
int dky8[] = {2, 2, -2, -2, 1, -1, 1, -1};
int tc = 1;
const double eps = 1e-9;
const double pi = acos(-1.0);
const long long int mx = 1e5;
const long long int mod = 1e9 + 7;
int main() {
  double w, h, a;
  scanf("%lf %lf %lf", &w, &h, &a);
  if (fabs(a - 90) < 1e-6) {
    printf("%.10lf\n", min(w, h) * min(w, h));
    return 0;
  }
  if (a > 90) a = 180 - a;
  if (fabs(a) < 1e-6) {
    printf("%.10lf\n", w * h);
    return 0;
  }
  a = a / 180 * pi;
  if (w < h) swap(w, h);
  double ans = w * h;
  double p1x = -w / 2 * cos(a) - h / 2 * sin(a);
  double p1y = -w / 2 * sin(a) + h / 2 * cos(a);
  double p2x = w / 2 * cos(a) - h / 2 * sin(a);
  double p2y = w / 2 * sin(a) + h / 2 * cos(a);
  double p3x = -w / 2 * cos(a) + h / 2 * sin(a);
  double p3y = -w / 2 * sin(a) - h / 2 * cos(a);
  if (a <= atan2(h, w) * 2) {
    double xi1 =
        ((p2y - h / 2) * p1x + (h / 2 - p1y) * p2x) / (p2y - p1y) + w / 2;
    double yi1 =
        h / 2 - ((p2x - (-w / 2)) * p1y + ((-w / 2) - p1x) * p2y) / (p2x - p1x);
    double xi2 =
        ((p3y - (-h / 2)) * p1x + ((-h / 2) - p1y) * p3x) / (p3y - p1y) + w / 2;
    double yi2 =
        ((p3x - (-w / 2)) * p1y + ((-w / 2) - p1x) * p3y) / (p3x - p1x) -
        (-h / 2);
    ans -= xi1 * yi1 + xi2 * yi2;
  } else {
    double xi1 =
        ((p2y - h / 2) * p1x + (h / 2 - p1y) * p2x) / (p2y - p1y) + w / 2;
    double xi2 =
        ((p2y - (-h / 2)) * p1x + ((-h / 2) - p1y) * p2x) / (p2y - p1y) + w / 2;
    ans -= h * (xi1 + xi2);
  }
  printf("%.10lf\n", ans);
  return 0;
}