Codeforces 140A New Year Table

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.

Example 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:

ummmm..一道数学题，题意大概是一个大圆桌能放几个小圆，而且小圆要紧贴大圆边界。画图发现可以求出每个小圆的圆心角，比较圆心角的合与2*PI即可。但是这个题还要考虑精度的问题orz....导致一直WA 最后参考了一下别人的代码。 TIP: PI= acos(-1.0) #include <cstdio> #include <iostream> #include <cmath> using namespace std; #define PI acos(-1.0) int main(int argc, char const *argv[]) { int n,R,r; cin>>n>>R>>r; if(r>R) { cout<<"NO"<<endl; return 0; } if(r>(R-r)) { if(n==1) cout<<"YES"<<endl; else cout<<"NO"<<endl; return 0; } double agl=asin((double)r/(double)(R-r)); if(2.0*PI-2.0*agl*n>=-1e-12) cout<<"YES"<<endl; else cout<<"NO"<<endl; return 0; }