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 N;
char s[104];

int main(){
    scanf("%d %s", &N, &s);
    int cnt1 = 0, cnt2 = 0;
    for(int i=0;i<N;++i) s[i]=='R'? ++cnt1:++cnt2;
    printf("%s\n", cnt1>cnt2? "Yes":"No");
    return 0;
}
