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;
int main(){
  int n;
  cin >> n;
  char s[n+1];
  cin >> s;
  int c=0;
  for(int i=0;i<n;i++){
    if(s[i]=='B')c--;
    else c++;
  }
  if(c>0)cout << "Yes";
  else cout << "No";
}
