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.

<image>

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;
inline long long getint() {
  long long _x = 0, _tmp = 1;
  char _tc = getchar();
  while ((_tc < '0' || _tc > '9') && _tc != '-') _tc = getchar();
  if (_tc == '-') _tc = getchar(), _tmp = -1;
  while (_tc >= '0' && _tc <= '9') _x *= 10, _x += (_tc - '0'), _tc = getchar();
  return _x * _tmp;
}
long long mypow(long long _a, long long _x, long long _mod) {
  if (_x == 0) return 1ll;
  long long _tmp = mypow(_a, _x / 2, _mod);
  _tmp = (_tmp * _tmp) % _mod;
  if (_x & 1) _tmp = (_tmp * _a) % _mod;
  return _tmp;
}
inline long long add(long long _x, long long _y,
                     long long _mod = 1000000007ll) {
  long long _ = _x + _y;
  if (_ >= _mod) _ -= _mod;
  return _;
}
inline long long sub(long long _x, long long _y,
                     long long _mod = 1000000007ll) {
  long long _ = _x - _y;
  if (_ < 0) _ += _mod;
  return _;
}
inline long long mul(long long _x, long long _y,
                     long long _mod = 1000000007ll) {
  long long _ = _x * _y;
  if (_ >= _mod) _ %= _mod;
  return _;
}
inline bool equal(double _x, double _y) {
  return _x > _y - 1e-9 && _x < _y + 1e-9;
}
int __ = 1, _cs;
void build() {}
void Print(const pair<long double, long double>& tp) {
  printf("%.6f %.6f\n", (double)tp.first, (double)tp.second);
}
inline pair<long double, long double> operator+(
    const pair<long double, long double>& p1,
    const pair<long double, long double>& p2) {
  return make_pair(p1.first + p2.first, p1.second + p2.second);
}
inline pair<long double, long double> operator-(
    const pair<long double, long double>& p1,
    const pair<long double, long double>& p2) {
  return make_pair(p1.first - p2.first, p1.second - p2.second);
}
inline pair<long double, long double> operator*(
    const long double& tk, const pair<long double, long double>& tp) {
  return make_pair(tk * tp.first, tk * tp.second);
}
inline long double operator^(const pair<long double, long double>& p1,
                             const pair<long double, long double>& p2) {
  return p1.first * p2.second - p1.second * p2.first;
}
pair<long double, long double> inter(pair<long double, long double> p1,
                                     pair<long double, long double> p2,
                                     pair<long double, long double> q1,
                                     pair<long double, long double> q2) {
  long double f1 = (p2 - p1) ^ (q1 - p1);
  long double f2 = (p2 - p1) ^ (p1 - q2);
  long double f = (f1 + f2);
  return (f2 / f) * q1 + (f1 / f) * q2;
}
long long n;
long double r;
pair<long double, long double> p1, p2, p3, p4;
const long double PI = acosl(-1.0);
void init() {
  n = getint();
  r = getint();
  p1 = make_pair(cosl(((long double)(n / 2) / (long double)n) * 2.0 * PI),
                 sinl(((long double)(n / 2) / (long double)n) * 2.0 * PI));
  p2 = make_pair(cosl(((long double)(n / 2 + 1) / (long double)n) * 2.0 * PI),
                 sinl(((long double)(n / 2 + 1) / (long double)n) * 2.0 * PI));
  p3 = make_pair(cosl(((long double)(n - 1) / (long double)n) * 2.0 * PI),
                 sinl(((long double)(n - 1) / (long double)n) * 2.0 * PI));
  p4 = make_pair(cosl((1.0 / (long double)n) * 2.0 * PI),
                 sinl((1.0 / (long double)n) * 2.0 * PI));
}
void solve() {
  pair<long double, long double> mid = inter(p1, p3, p2, p4);
  long double area = (p1 ^ mid) + (mid ^ p2);
  area *= (long double)n * r * r * 0.5;
  if (area < 0) area = -area;
  printf("%.12f\n", (double)area);
}
int main() {
  build();
  while (__--) {
    init();
    solve();
  }
}