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;
string s;
cin >> N >> s;
int r = 0;
int b = 0;
for(int i=0;i<N;i++){
if(s.at(i) == 'R') r++;
else b++;
}
if(r > b) cout << "Yes" <<endl;
else cout << "No" << endl;
}
