New Year Table（几何）

New Year Table

Time Limit:2000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 140A

Description

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.

Sample Input

Input

4 10 4

Output

YES

Input

5 10 4

Output

NO

Input

1 10 10

Output

YES

Hint

The possible arrangement of the plates for the first sample is

题解：小圆贴着大圆的边，问是否能放n个；注意精度；

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#in\
clude<algorithm>
using namespace std;
const double PI = acos(-1.0);
int main(){
    int n, R, r;
    while(~scanf("%d%d%d", &n, &R, &r)){
        if(r > R){
            if(n == )
                puts("YES");
            else
                puts("NO");
            continue;
        }
        else if(r == R){
            if(n <= )
                puts("YES");
            else
                puts("NO");
            continue;
        }

        else if( * r > R){
            if(n <= )
                puts("YES");
            else
                puts("NO");
            continue;
        }
        double cosa = (2.0 * (R - r) * (R - r) - (2.0*r) * (2.0*r))/ (2.0*(R-r)*(R-r));
        double a = acos(cosa);
        //printf("%lf\n", 2.0 * PI / a);
        if((n - 2.0 * PI / a) <= 1e-){
            puts("YES");
        }
        else{
            puts("NO");
        }
    }
}