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 Python (3.4.3) (judge verdict: Wrong Answer)

# coding: utf-8

import sys
import math

import array
import bisect
import collections
from collections import Counter, defaultdict
import fractions
import heapq
import re

sys.setrecursionlimit(1000000)


def array2d(dim1, dim2, init=None):
    return [[init for _ in range(dim2)] for _ in range(dim1)]

def argsort(l, reverse=False):
    return sorted(range(len(l)), key=lambda i: l[i], reverse=reverse)

def argmin(l):
    return l.index(min(l))

def YESNO(ans, yes="YES", no="NO"):
    print([no, yes][ans])

II = lambda: int(input())
MI = lambda: map(int, input().split())
MIL = lambda: list(MI())
MIS = lambda: input().split()


def main():
    N = II()
    c = collections.Counter(input())
    return c.most_common()[0][0] == "R"


if __name__ == "__main__":
    YESNO(main(), "Yes", "No")