Codeforces-----140A---New Year Table---数学水题

探讨如何在圆形晚宴桌上放置多个等大小的小圆盘，确保每个圆盘都紧贴桌子边缘且不互相重叠的技术挑战。通过数学计算确定桌子是否足够大以容纳指定数量的圆盘。

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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:

$\text{[math]}$

水题，问一个大圆里能不能有n个等大的小圆，小圆必须贴着大圆的外围放，n<3的时候很明显
n>3的时候可以由上图得出，连接小圆圆心，是一个正n边形，可以求出大圆的半径就是小圆半径加上正n边形内心到顶点距离
正n边形可以分成n个等大的三角形，内角可以求出，可由三角函数求出距离，c++的三角函数是求的弧度，需要转化一下2*π = 360°

#include<cstdio>
#include<algorithm>
#include<cmath>
#include<string>
#include<cstring>
#include<vector>
#include<map>
#include<queue>
#include<iostream>
#define PI acos(-1.0)
#define inf 0x3f3f3f3f
using namespace std; 
const int maxn = 1e5+10;
typedef long long LL;
using namespace std;
int main(){
	int t, m, n, ans;
	cin >> t >> m >> n;
	if(t == 1){
		if(n > m) cout << "NO" << endl;
		else cout << "YES" << endl;
	}
	else if(t == 2){
		if(n*2 > m) cout << "NO" << endl;
		else cout << "YES" << endl;
	}
	else{
		double c = 360.0/t;
		double a, b;
		a = sin(c*2*PI/360);
		c = (180-c)/2;
		b = sin(c*2*PI/360);
		double x = 2*n*b/a;
		if(x+n > m) cout << "NO" << endl;
		else cout << "YES" << endl;
	}
	
	return 0;
}