Problem:
Score : 200 points

Problem Statement

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.

Sample Input 1

4
RRBR

Sample Output 1

Yes

There are three people wearing a red hat, and one person wearing a blue hat.

Since there are more people wearing a red hat than people wearing a blue hat, the answer is Yes.

Sample Input 2

4
BRBR

Sample Output 2

No

There are two people wearing a red hat, and two people wearing a blue hat.

Since there are as many people wearing a red hat as people wearing a blue hat, the answer is No.

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <vector>
#define llong long long
#define INF (__INT32_MAX__ / 2)
#define MAX_N 100

using namespace std;

int main(int argc, char **argv) {
    int n;
    char str[MAX_N + 1];
    scanf("%d", &n);
    scanf("%s", str);

    int red = 0, blue = 0;
    for(int i = 0; i < n; ++i)
        if(str[i] == 'R')
            ++red;
        else if(str[i] == 'B')
            ++blue;
    
    printf(red > blue ? "Yes\n" : "No\n");
}