Problem: You are given two circles. Find the area of their intersection.

Input

The first line contains three integers x1, y1, r1 ( - 109 ≤ x1, y1 ≤ 109, 1 ≤ r1 ≤ 109) — the position of the center and the radius of the first circle.

The second line contains three integers x2, y2, r2 ( - 109 ≤ x2, y2 ≤ 109, 1 ≤ r2 ≤ 109) — the position of the center and the radius of the second circle.

Output

Print the area of the intersection of the circles. The answer will be considered correct if the absolute or relative error doesn't exceed 10 - 6.

Examples

Input

0 0 4
6 0 4


Output

7.25298806364175601379


Input

0 0 5
11 0 5


Output

0.00000000000000000000

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long double x[2], y[2], r[2];
inline long double sqr(long double x) { return x * x; }
inline long long sqrl(long long x) { return x * x; }
inline long double fcos(long double a, long double b, long double c) {
  return acosl((a * a + b * b - c * c) / (2 * a * b));
}
inline long double cut(long double ang, long double r) {
  long double s1 = ang * r * r;
  long double s2 = sinl(ang) * cosl(ang) * r * r;
  return s1 - s2;
}
const long double pi = acosl(-1);
double solve() {
  if (r[0] > r[1]) {
    swap(r[0], r[1]);
    swap(x[0], x[1]);
    swap(y[0], y[1]);
  }
  long long dx = x[0] - x[1], dy = y[0] - y[1];
  long long ds = sqrl(dx) + sqrl(dy);
  if (ds >= sqrl(r[0] + r[1])) return 0;
  long double d = sqrtl(ds);
  if (d + r[0] <= r[1]) return sqr(r[0]) * pi;
  long double ang[2];
  ang[0] = fcos(d, r[0], r[1]);
  ang[1] = fcos(d, r[1], r[0]);
  return cut(ang[0], r[0]) + cut(ang[1], r[1]);
}
int main() {
  cin >> x[0] >> y[0] >> r[0] >> x[1] >> y[1] >> r[1];
  printf("%.10f\n", solve());
  return 0;
}