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 <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <utility>
#include <cstring>
#include <iomanip>
#include <numeric>
#include <cmath>
#include <cassert>
using namespace std;
using ll = long long;
const int INF = 1<<30;
const int MOD = 1e9 + 7;
signed main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n; cin >> n;
    string s; cin >> s;
    int R = 0, B = 0;
    for(int i = 0; i < s.size(); i++)
    {
        if(s[i] == 'R') R++;
        else B++;
    }
    cout << (R > B ? "Yes" : "No") << endl;
    return 0;
}