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<string>
#include<iostream>
using namespace std;

int main(){
int N; string s;
cin >> N;
cin >> s;
int k = 0;
for(int j = 0; j < N; j++){
if(s[j] == 'R'){
k++;
}
}
if(k > N - k){
cout << "Yes" << endl;
}
else{
cout << "No" << endl;
}
return 0;
}