2 seconds

256 megabytes

standard input

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.

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.

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.

/* ***********************************************

Author        :guanjun

Created Time  :2016/7/26 11:34:41

File Name     :cf100a.cpp

************************************************ */

#include <iostream>

#include <cstring>

#include <cstdlib>

#include <stdio.h>

#include <algorithm>

#include <vector>

#include <queue>

#include <set>

#include <map>

#include <string>

#include <math.h>

#include <stdlib.h>

#include <iomanip>

#include <list>

#include <deque>

#include <stack>

#define ull unsigned long long

#define ll long long

#define mod 90001

#define INF 0x3f3f3f3f

#define maxn 10010

#define cle(a) memset(a,0,sizeof(a))

const ull inf = 1LL << ;

const double eps=1e-;

#define pi 4.0*atan(1.0)

using namespace std;

priority_queue<int,vector<int>,greater<int> >pq;

struct Node{

    int x,y;

};

struct cmp{

    bool operator()(Node a,Node b){

        if(a.x==b.x) return a.y> b.y;

        return a.x>b.x;

    }

};

bool cmp(int a,int b){

    return a>b;

}

int main()

{

    #ifndef ONLINE_JUDGE

    //freopen("in.txt","r",stdin);

    #endif

    //freopen("out.txt","w",stdout);

    double n,R,r;

    while(cin>>n>>R>>r){

        if(n==){

            if(r<=R)puts("YES");

            else puts("NO");

            continue;

        }

        double tmp=(pi/n);

        double c=sin(tmp)*(R-r);

        if(r<=c+eps){

            puts("YES");

        }

        else puts("NO");

    }

    return ;

}