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>
#pragma GCC optimize("O3")
using namespace std;
const int INF = 1000 * 1000 * 1000 + 322;
const long long LLINF =
    2LL * 1000LL * 1000LL * 1000LL * 1000LL * 1000LL * 1000LL + 57;
const long long MOD = 1000 * 1000 * 1000 + 7;
const int RMOD = MOD - 2;
const int N = 200 * 1000 + 57;
const int P1M = 1336337;
const int P1K = 1009;
const long double EPS = 1e-9;
const long double PI = 3.1415926535897932384626433832795;
template <typename T, typename U>
inline ostream &operator<<(ostream &_out, const pair<T, U> &_p) {
  _out << _p.x << " " << _p.y;
  return _out;
}
template <typename T, typename U>
inline istream &operator>>(istream &_in, pair<T, U> &_p) {
  _in >> _p.x >> _p.y;
  return _in;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const vector<T> &_v) {
  if (_v.empty()) return _out;
  _out << _v.front();
  for (auto _it = ++_v.begin(); _it != _v.end(); ++_it) _out << ' ' << *_it;
  return _out;
}
template <typename T>
inline istream &operator>>(istream &_in, vector<T> &_v) {
  for (auto &_i : _v) _in >> _i;
  return _in;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const set<T> &_s) {
  if (_s.empty()) return _out;
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) _out << ' ' << *_it;
  return _out;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const multiset<T> &_s) {
  if (_s.empty()) return _out;
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) _out << ' ' << *_it;
  return _out;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const unordered_set<T> &_s) {
  if (_s.empty()) return _out;
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) _out << ' ' << *_it;
  return _out;
}
template <typename T>
inline ostream &operator<<(ostream &_out, const unordered_multiset<T> &_s) {
  if (_s.empty()) return _out;
  _out << *_s.begin();
  for (auto _it = ++_s.begin(); _it != _s.end(); ++_it) _out << ' ' << *_it;
  return _out;
}
template <typename T, typename U>
inline ostream &operator<<(ostream &_out, const map<T, U> &_m) {
  if (_m.empty()) return _out;
  _out << '(' << _m.begin()->x << ": " << _m.begin()->y << ')';
  for (auto _it = ++_m.begin(); _it != _m.end(); ++_it)
    _out << ", (" << _it->x << ": " << _it->y << ')';
  return _out;
}
template <typename T, typename U>
inline ostream &operator<<(ostream &_out, const unordered_map<T, U> &_m) {
  if (_m.empty()) return _out;
  _out << '(' << _m.begin()->x << ": " << _m.begin()->y << ')';
  for (auto _it = ++_m.begin(); _it != _m.end(); ++_it)
    _out << ", (" << _it->x << ": " << _it->y << ')';
  return _out;
}
const char *infile = "";
const char *outfile = "";
inline long long gcd(long long a, long long b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
inline long long pwm(long long xx, long long pow, long long MD = MOD) {
  long long mlt = 1;
  while (pow) {
    if (pow & 1) {
      mlt *= xx;
      xx *= xx;
      pow >>= 1;
      xx %= MD;
      mlt %= MD;
    } else {
      pow >>= 1;
      xx *= xx;
      xx %= MD;
    }
  }
  return mlt;
}
inline long long pw(long long xx, int pow) {
  long long mlt = 1;
  while (pow) {
    if (pow & 1) {
      mlt *= xx;
      xx *= xx;
      pow >>= 1;
    } else {
      pow >>= 1;
      xx *= xx;
    }
  }
  return mlt;
}
inline long long inv(long long r) { return pwm(r, RMOD); }
long double seg(long double n, long double r) {
  long double si1 = sin(2 * PI / n);
  long double co2 = cos((PI - 3 * PI / n) / 2);
  long double co3 = cos((PI - 2 * PI / n) / 2);
  long double l1 = r * co3;
  long double l2 = l1 / co2;
  long double p = l1 + l2, pp = l2 - l1;
  long double striang = sqrt(p * l1 * pp * l1);
  return striang + r * r / 2 * (2 * PI / n - si1);
}
inline void solve(long double tt) {
  long double n, r;
  cin >> n >> r;
  long double s = PI * r * r - n * seg(n, r);
  cout << s << "\n";
}
int main() {
  long double tt = clock();
  if (*infile != '\0') freopen(infile, "r", stdin);
  if (*outfile != '\0') freopen(outfile, "w", stdout);
  cout.tie(nullptr);
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  cout << setprecision(20);
  cout << fixed;
  solve(tt);
  return 0;
}