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;
typedef long long ll;


int main(){
	ll n;
	string s;
	cin>>n>>s;
	ll bc=0,rc=0;
	for(int i=0;i<s.size();i++){
		if(s[i]=='B')bc++;
		else rc++;
	}
	cout<<(rc>bc?"Yes":"No");
	return 0;
}
