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;
const double eps = 1e-8;
const double pi = acos(-1.0);
int sign(double k) {
  if (k > eps)
    return 1;
  else if (k < -eps)
    return -1;
  return 0;
}
int cmp(double k1, double k2) { return sign(k1 - k2); }
struct point {
  double x, y, r;
  point operator+(const point &k1) const { return (point){k1.x + x, k1.y + y}; }
  point operator-(const point &k1) const { return (point){x - k1.x, y - k1.y}; }
  point operator*(double k1) const { return (point){x * k1, y * k1}; }
  point operator/(double k1) const { return (point){x / k1, y / k1}; }
  int operator==(const point &k1) const {
    return cmp(x, k1.x) == 0 && cmp(y, k1.y) == 0;
  }
  bool operator<(const point k1) const {
    int a = cmp(x, k1.x);
    if (a == -1)
      return 1;
    else if (a == 1)
      return 0;
    else
      return cmp(y, k1.y) == -1;
  }
};
double dis(point xx, point yy) {
  return sqrt((yy.x - xx.x) * (yy.x - xx.x) + (yy.y - xx.y) * (yy.y - xx.y));
}
double abs1(point a) { return sqrt(a.x * a.x + a.y * a.y); }
double abs2(point a) { return a.x * a.x + a.y * a.y; }
double dot(point k1, point k2) { return k1.x * k2.x + k1.y * k2.y; }
double cross(point k1, point k2) { return k1.x * k2.y - k1.y * k2.x; }
vector<point> ConvexHull(vector<point> A, int flag = 1) {
  int n = A.size();
  vector<point> ans(n * 2);
  sort(A.begin(), A.end());
  int now = -1;
  for (int i = 0; i < A.size(); i++) {
    while (now > 0 &&
           sign(cross(ans[now] - ans[now - 1], A[i] - ans[now - 1])) < flag)
      now--;
    ans[++now] = A[i];
  }
  for (int i = n - 2, pre = now; i >= 0; i--) {
    while (now > pre &&
           sign(cross(ans[now] - ans[now - 1], A[i] - ans[now - 1])) < flag)
      now--;
    ans[++now] = A[i];
  }
  ans.resize(now);
  return ans;
}
int main() {
  double w, h, af;
  cin >> w >> h >> af;
  if (w < h) swap(w, h);
  af = min(af, 180 - af);
  af = af / 180 * pi;
  double ans = w * h;
  double h1 = w / 2 - tan(af / 2) * h / 2;
  double h2 = h / 2 - tan(af) * h1 + tan(af / 2) * (w / 2);
  double b = 2 * atan(h / w);
  if (af < b) {
    if (af > 0) ans -= (h1 * h1 * tan(af) + h2 * h2 * sin(af) * cos(af));
  } else {
    ans = h * h / sin(af);
  }
  printf("%.10lf\n", ans);
}