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,red=0; cin>>n;
  for(int i=0;i<n;i++){
    char a; cin>>a;
    if(a=='R')red++;
  }
  
  cout<<(red > n-red ? "Yes" : "No")<<endl;
  return 0;
}