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`.

Examples

Input

4
RRBR


Output

Yes


Input

4
BRBR


Output

No
import java.io.*;
import java.util.*;
public class Main{
    public static void main(String[] args) throws IOException{
        Reader r = new Reader();  
        int N = r.i();
        String s = r.s();
        int rn = 0;
        for(int i = 0; i < N; i++)
            if(s.charAt(i) == 'R')
                rn++;
        println(N/2 < rn ? "Yes" : "No");
    }
    static void print(Object o){
        System.out.print(o.toString());
    }
    static void println(Object o){
        System.out.println(o.toString());
    }
    static int Int(String s){
        return Integer.parseInt(s);
    }
    static long Long(String s){
        return Long.parseLong(s);
    }
    static class Reader extends BufferedReader{
        Reader(){
            super(new InputStreamReader(System.in));
        }
        String s() throws IOException{
            return readLine();
        }
        String[] ss() throws IOException{
            return s().split(" ");
        }
        int i() throws IOException{
            return Int(s());
        }
        int[] ii() throws IOException{
            String[] ss = ss();
            int size = ss.length;
            int[] ii = new int[size];
            for(int j = 0; j < size; j++) ii[j] = Int(ss[j]);
            return ii;
        }
        long l() throws IOException{
            return Long(s());
        }
        long[] ll() throws IOException{
            String[] ss = ss();
            int size = ss.length;
            long[] ll = new long[size];
            for(int j = 0; j < size; j++) ll[j] = Long(ss[j]);            
            return ll;
        }
    }
}
