Problem: Problem

Find the area of ​​a regular N / K polygon inscribed in a circle with a radius of 1.

However, a regular N / K polygon is defined as "the outermost figure that takes N points on the circumference at equal intervals and connects each point every K-1".

For example, a 5/2 polygon can be drawn as follows. First, take five points at equal intervals on the circumference of radius 1.

Sample Input 2 diagram


Next, connect each point every other 2-1 = 1.

Sample Input 2 diagram


The outermost figure is a regular 5/2 square.

Sample Input 2 diagram

Constraints

The input satisfies the following constraints.

* 5 ≤ N ≤ 106
* 1 <K <N / 2
* N and K are integers that are relatively prime

Input

The input is given in the following format.


N K


Two integers N and K are given on one line.

Output

Output the area of ​​a regular N / K polygon inscribed in a circle with a radius of 1 on one line. An error of 10-5 or less is acceptable.

Examples

Input

5 2


Output

1.12256994


Input

20 3


Output

2.93114293


Input

7 3


Output

1.08395920


Input

100000 3


Output

3.14159265

Solution (in CPP):

Solution:
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>

#include <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <queue>
#include <stack>
#include <map>
#include <set>

#include <functional>
#include <cassert>

typedef long long ll;
using namespace std;

#define debug(x) cerr << #x << " = " << x << endl;


#define mod 1000000007 //1e9+7(prime number)
#define INF 1000000000 //1e9
#define LLINF 2000000000000000000LL //2e18
#define SIZE 10000


int main(){
  int n,k;
  double ans;
  
  cin >> n >> k;

  ans = 1 * 1 * sin(2*M_PI/n) * n/2;
  
  if(k > 1){
    double a = sqrt(2 * (1 - cos(2*M_PI/n)));
    ans -= (tan(M_PI/n*(k-1)) * a / 2) * a / 2 * n;
  }
  
  printf("%.10lf\n",ans);
  
  return 0;
}