We have a special convex that all points have the same distance to origin point.
As you know we can get N segments after linking the origin point and the points on the convex. We can also get N angles between each pair of the neighbor segments.
Now give you the data about the angle, please calculate the area of the convex

Input

There are multiple test cases.
The first line contains two integer N and D indicating the number of the points and their distance to origin. (3 <= N <= 10, 1 <= D <= 10)
The next lines contain N integers indicating the angles. The sum of the N numbers is always 360.

Output

For each test case output one float numbers indicating the area of the convex. The printed values should have 3 digits after the decimal point.
Sample Input

4 1
90 90 90 90
6 1
60 60 60 60 60 60
Sample Output

2.000
2.598

n邊形的面積可以拆分成n個三角形面積之和

而已知每個三角形的兩邊及兩邊夾角，我們可以通過三角形面積公式算出每個三角形的面積，相加之和便是最終的n邊形面積

水題，直接求所有三角形的面積即可 一開始想複雜了，即使是有角度大於180的也不需要分開考慮，應為此時sin為負就相當於對面積進行減了

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
int main()
{
    int n,d,i,x;
    double ans;
    while(~scanf("%d%d",&n,&d))
    {
        ans=0;
        for(i=0;i<n;i++)
        {
            scanf("%d",&x);
            ans+=d*d*sin(PI*x/180)/2;
        }
        printf("%.3f\n",ans);
    }
    return 0;
}