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;
clock_t __START, __END;
double __TOTALTIME;
void _MS() { __START = clock(); }
void _ME() {
  __END = clock();
  __TOTALTIME = (double)(__END - __START) / CLOCKS_PER_SEC;
  cout << "Time: " << __TOTALTIME << " s" << endl;
}
const double E = 2.718281828;
const double PI = acos(-1.0);
const long long INF = (1LL << 60);
const int inf = (1 << 30);
const double ESP = 1e-9;
const int mod = (int)1e9 + 7;
const int N = (int)1e6 + 10;
bool zero(double x) { return (x > 0 ? x : -x) < ESP; }
struct point {
  double x, y;
};
struct line {
  point a, b;
};
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);
}
double dmult(point p1, point p2, point p0) {
  return (p1.x - p0.x) * (p2.x - p0.x) + (p1.y - p0.y) * (p2.y - p0.y);
}
double distance(point p1, point p2) {
  return sqrt((p1.x - p2.x) * (p1.x - p2.x) + (p1.y - p2.y) * (p1.y - p2.y));
}
bool isIntersected(point s1, point e1, point s2, point e2) {
  return (max(s1.x, e1.x) >= min(s2.x, e2.x)) &&
         (max(s2.x, e2.x) >= min(s1.x, e1.x)) &&
         (max(s1.y, e1.y) >= min(s2.y, e2.y)) &&
         (max(s2.y, e2.y) >= min(s1.y, e1.y)) &&
         (xmult(s1, s2, e1) * xmult(s1, e1, e2) > 0) &&
         (xmult(s2, s1, e2) * xmult(s2, e2, e1) > 0);
}
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;
}
point rotate(point p0, double alpha) {
  alpha = alpha / 180 * PI;
  return {p0.x * cos(alpha) - p0.y * sin(alpha),
          p0.y * cos(alpha) + p0.x * sin(alpha)};
}
point Pin[20];
int cmp(point a, point b) {
  if (xmult(a, b, Pin[1]) > ESP) return 1;
  if (fabs(xmult(a, b, Pin[1])) < ESP &&
      distance(b, Pin[1]) - distance(a, Pin[1]) > ESP)
    return 1;
  return 0;
}
point Stack[N];
int top;
void Graham(int n, point p[]) {
  top = 3;
  sort(p + 2, p + 1 + n, cmp);
  Stack[1] = p[1];
  Stack[2] = p[2];
  Stack[3] = p[3];
  for (int i = 4; i <= n; i++) {
    while (top >= 1 && xmult(p[i], Stack[top], Stack[top - 1]) > ESP) {
      top--;
    }
    Stack[++top] = p[i];
  }
}
double getArea() {
  double sum = fabs(xmult(Stack[2], Stack[3], Stack[1]));
  for (int i = 3; i < top; i++)
    sum += fabs(xmult(Stack[i], Stack[i + 1], Stack[1]));
  return sum / 2.0;
}
int main() {
  double w, h, alpha;
  scanf("%lf%lf%lf", &w, &h, &alpha);
  if (alpha == 0 || alpha == 180) {
    printf("%.10lf\n", w * h);
    return 0;
  }
  if (alpha == 90 && w == h) {
    printf("%.10lf\n", w * h);
    return 0;
  }
  line line1[10];
  point p1 = {-w / 2, h / 2};
  point p2 = {w / 2, h / 2};
  point p3 = {w / 2, -h / 2};
  point p4 = {-w / 2, -h / 2};
  line1[1] = {p1, p2};
  line1[2] = {p2, p3};
  line1[3] = {p3, p4};
  line1[4] = {p4, p1};
  line line2[10];
  p1 = rotate(p1, alpha);
  p2 = rotate(p2, alpha);
  p3 = rotate(p3, alpha);
  p4 = rotate(p4, alpha);
  line2[1] = {p1, p2};
  line2[2] = {p2, p3};
  line2[3] = {p3, p4};
  line2[4] = {p4, p1};
  int cp = 0;
  for (int i = 1; i <= 4; ++i) {
    for (int j = 1; j <= 4; ++j) {
      if (isIntersected(line1[i].a, line1[i].b, line2[j].a, line2[j].b))
        Pin[++cp] =
            intersection(line1[i].a, line1[i].b, line2[j].a, line2[j].b);
    }
  }
  Graham(cp, Pin);
  printf("%.10lf\n", getArea());
  return 0;
}