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;
typedef priority_queue<long long, vector<long long>, greater<long long>>
    minHeap;
const long long MD = 1e9 + 7;
inline long long mul(long long a, long long b) { return (a * b) % MD; }
inline long long po(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b & 1) ans = mul(ans, a);
    a = mul(a, a);
    b /= 2;
  }
  return ans;
}
long long INF = 1e17;
const int N = 2e5 + 22;
long long n, m, t, k;
long long arr[N];
long long x;
string s;
long double x1, x2, y11, y2, r1, r2;
long double doo = 2;
long double pi = doo * asin(1);
int main() {
  ios::sync_with_stdio(false);
  cin >> x1 >> y11 >> r1 >> x2 >> y2 >> r2;
  long double d = sqrt((x1 - x2) * (x1 - x2) + (y11 - y2) * (y11 - y2));
  if (d >= r1 + r2) {
    printf("%.7lf", 0);
    return 0;
  } else if (d <= abs(r1 - r2)) {
    long double r = min(r1, r2);
    printf("%.7lf", double(pi * r * r));
    return 0;
  }
  long double teta;
  teta = acos((d * d + r1 * r1 - r2 * r2) / (doo * d * r1));
  long double ans = 0;
  ans += teta * (r1 * r1);
  ans -= r1 * r1 * sin(doo * teta) / doo;
  teta = acos((d * d - r1 * r1 + r2 * r2) / (doo * d * r2));
  ans += teta * (r2 * r2);
  ans -= r2 * r2 * sin(doo * teta) / doo;
  printf("%.7lf\n", (double)ans);
  return 0;
}