cf - 140 A. New Year Table（模拟）

A. New Year Table

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Gerald is setting the New Year table. The table has the form of a circle; its radius equals R. Gerald invited many guests and is concerned whether the table has enough space for plates for all those guests. Consider all plates to be round and have the same radii that equal r. Each plate must be completely inside the table and must touch the edge of the table. Of course, the plates must not intersect, but they can touch each other. Help Gerald determine whether the table is large enough for n plates.

Input

The first line contains three integers n, R and r (1 ≤ n ≤ 100, 1 ≤ r, R ≤ 1000) — the number of plates, the radius of the table and the plates' radius.

Output

Print "YES" (without the quotes) if it is possible to place n plates on the table by the rules given above. If it is impossible, print "NO".

Remember, that each plate must touch the edge of the table.

Examples

input

4 10 4

output

YES

input

5 10 4

output

NO

input

1 10 10

output

YES

Note

The possible arrangement of the plates for the first sample is:

 

题意：给一个大圆和一个小圆的半径，小圆必须在大圆内且与大圆相切。求满足该要求的小圆最多放多少个。思路：根据相切可以找到边角关系。小圆圆心到切点为（r），小圆圆心到大圆圆心为（R-r）。则一个圆占得圆心角为 2*arcsin(r/(R*r))。用2*PI 除以这个圆心角就可以得到最大个数。这题会卡精度问题，wa了N久。以后判断精度在结果加上 1e-12 后进行判断。代码：

#include <bits/stdc++.h>
using namespace std;

int main()
{
    double n,R,r;
    while(cin>>n>>R>>r){
        if(r > R){
            cout<<"NO"<<endl;
            continue;
        }
        if(r > R/2){
            if(n == 1)  cout<<"YES"<<endl;
            else
                cout<<"NO"<<endl;
            continue;
        }

        double a = asin((double)r/(R-r));

        if(3.1415926535898/a > n)
            cout<<"YES"<<endl;
        else
            cout<<"NO"<<endl;
    }
    return 0;
}