题目链接

NN is an experienced internet user and that means he spends a lot of time on the social media. Once he found the following image on the Net, which asked him to compare the sizes of inner circles:

It turned out that the circles are equal. NN was very surprised by this fact, so he decided to create a similar picture himself.

He managed to calculate the number of outer circles nn and the radius of the inner circle rr. NN thinks that, using this information, you can exactly determine the radius of the outer circles RR so that the inner circle touches all of the outer ones externally and each pair of neighboring outer circles also touches each other. While NN tried very hard to guess the required radius, he didn't manage to do that.

Help NN find the required radius for building the required picture.

Input

The first and the only line of the input file contains two numbers nn and rr (3≤n≤1003≤n≤100, 1≤r≤1001≤r≤100) — the number of the outer circles and the radius of the inner circle respectively.

Output

Output a single number RR — the radius of the outer circle required for building the required picture.

Your answer will be accepted if its relative or absolute error does not exceed 10−610−6.

Formally, if your answer is aa and the jury's answer is bb. Your answer is accepted if and only when |a−b|max(1,|b|)≤10−6|a−b|max(1,|b|)≤10−6.

Examples

input

3 1

output

6.4641016

input

6 1

output

1.0000000

input

100 100

output

3.2429391

PS：题意：就是给你n，r。n表示外面有多少个圆，r表示给出圆的半径。现在要求出满足上图摆设方式的小圆的半径。

题解：首先我们两个小园的圆心和大圆的圆心连接，就能得到如下图所示三角形。

我们设小圆的半径为R 我们可以得到左边（左右其实都一样）那个等腰三角形三角形的斜边长度为R+r，底边为R。又校园的所有圆心连接起来就是一个正多边形，我们知道多边形内角和为：π*（n-2）（这里外面小圆有多少个，n就为多少，这个可以在草稿本上画一下）。很明显，n个球可以分割成n个这样的等腰三角形。然后一个底角的角度为π*(n-2)/n/2;现在我们可以根据余弦公式得到：R/(R+r)=cos(a)；这样就可以推出R：R=R=r*cos(a)/(1-cos(a));

#include <iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<map>
#include<queue>
#include<set>
#include<cmath>
#include<stack>
#include<string>
const int maxn=1e5+10;
const int mod=1e9+7;
const int inf=1e8;
#define me(a,b) memset(a,b,sizeof(a))
#define lowbit(x) x&(-x)
#define mid (l+r)/2
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
#define PI 3.14159265358979323846
typedef long long ll;
using namespace std;
int main()
{
    double n,r;
    cin>>n>>r;
    double x=PI*(n-2)/n;
    printf("%.7f\n",r*cos(x/2)/(1-cos(x/2)));
    return 0;
}