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: Wrong Answer)

#include <cstdio>
#include <algorithm>    
#include <iostream>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <iomanip>
#include <cstring>
#include <cmath>
#define all(x) (x).begin(),(x).end()
using namespace std;
typedef long long LL;
typedef long double LD;
typedef unsigned int uii;
typedef pair<int,int> pii;
typedef unsigned long long uLL; 
typedef vector<int> vii; 
//__int128


int main(int argc, char const *argv[])
{
#ifndef ONLINE_JUDGE
	freopen("in.txt", "r", stdin);
	// freopen("out.txt", "w", stdout);
#endif
	int n;
	string s;
	int c = 0;
	cin >> n >> s;
	for(int i = 0; i < n; ++i){
		if(s[i] == 'R') 
			++c;
	}
	if(c > n - c){
		puts("Yes");
	} else {
		puts("No");
	}
	return 0;
}