Problem: It was decided in IT City to distinguish successes of local IT companies by awards in the form of stars covered with gold from one side. To order the stars it is necessary to estimate order cost that depends on the area of gold-plating. Write a program that can calculate the area of a star.

A "star" figure having n ≥ 5 corners where n is a prime number is constructed the following way. On the circle of radius r n points are selected so that the distances between the adjacent ones are equal. Then every point is connected by a segment with two maximally distant points. All areas bounded by the segments parts are the figure parts.

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Input

The only line of the input contains two integers n (5 ≤ n < 109, n is prime) and r (1 ≤ r ≤ 109) — the number of the star corners and the radius of the circumcircle correspondingly.

Output

Output one number — the star area. The relative error of your answer should not be greater than 10 - 7.

Examples

Input

7 10


Output

108.395919545675

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T1, class T2>
bool chmin(T1 &a, T2 b) {
  return b < a && (a = b, true);
}
template <class T1, class T2>
bool chmax(T1 &a, T2 b) {
  return a < b && (a = b, true);
}
const long double eps = 1e-8;
long double dot(complex<long double> a, complex<long double> b) {
  return real(conj(a) * b);
}
long double cross(complex<long double> a, complex<long double> b) {
  return imag(conj(a) * b);
}
int ccw(complex<long double> a, complex<long double> b,
        complex<long double> c) {
  b -= a;
  c -= a;
  if (cross(b, c) > eps) return 1;
  if (cross(b, c) < -eps) return -1;
  if (dot(b, c) < -eps) return 2;
  if (norm(b) < norm(c)) return -2;
  return 0;
}
bool intersectSS(complex<long double> p1, complex<long double> p2,
                 complex<long double> p3, complex<long double> p4) {
  return ccw(p1, p2, p3) * ccw(p1, p2, p4) <= 0 &&
         ccw(p3, p4, p1) * ccw(p3, p4, p2) <= 0;
}
complex<long double> intersection(complex<long double> p1,
                                  complex<long double> p2,
                                  complex<long double> q1,
                                  complex<long double> q2) {
  complex<long double> base = q2 - q1;
  long double d1 = abs(cross(base, p1 - q1));
  long double d2 = abs(cross(base, p2 - q1));
  return (p1 * d2 + p2 * d1) / (d1 + d2);
}
int main() {
  int n;
  long double r;
  cin >> n >> r;
  const long double pi = acos(-1);
  int d = (n - 1) / 2;
  long double theta = 2 * pi / n;
  complex<long double> p1 = polar(1.0l, theta * 0);
  complex<long double> p2 = polar(1.0l, theta * d);
  complex<long double> q1 = polar(1.0l, theta * 1);
  complex<long double> q2 = polar(1.0l, theta * (1 - d));
  complex<long double> p = intersection(p1, p2, q1, q2);
  long double area = abs(cross(p, p1)) * n * r * r;
  printf("%.20f\n", (double)area);
  return 0;
}