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 <bits/stdc++.h>
#define REP(i,n) for(ll i = 0; i < (ll)n; i++)
#define FOR(i,a,b) for (ll i=(a);i<(ll)b; i++)
#define ALL(obj) (obj).begin(),(obj).end()
#define INF 1000000000000000
using namespace std;
typedef long long ll;
typedef double db;
typedef string str;
typedef pair<ll,ll> p;
const int MOD = 1000000007;

int main(){
    int N;cin>>N;
    string S;cin>>S;
    int blue = 0;
    int red = 0;
    for(int i = 0; i<S.size();i++){
        if(S[i] == 'R'){
            red++;
        }else{
            blue++;
        }
    }
    if(red>blue){
        cout<<"Yes";
    }else{
        cout<<"No";
    }

}