There are N people numbered 1 to N. Each person wears a red hat or a blue hat.

You are given a string s representing the colors of the people. Person i wears a red hat if s_i is `R`, and a blue hat if s_i is `B`.

Determine if there are more people wearing a red hat than people wearing a blue hat.

Constraints

* 1 \leq N \leq 100
* |s| = N
* s_i is `R` or `B`.

Input

Input is given from Standard Input in the following format:


N
s


Output

If there are more people wearing a red hat than there are people wearing a blue hat, print `Yes`; otherwise, print `No`.

Examples

Input

4
RRBR


Output

Yes


Input

4
BRBR


Output

No
#include <bits/stdc++.h>
using namespace std;

char color[105];
int main(void){
  int N; cin>>N;
  cin>>color;
  int cnt=0;
  for(int i=0;i<N;i++){
    if(color[i]=='R') cnt++;
  }
  if(cnt*2>N) cout<<"Yes"<<endl;
  else cout<<"No"<<endl;
  return 0;
}
