An altar enshrines N stones arranged in a row from left to right. The color of the i-th stone from the left (1 \leq i \leq N) is given to you as a character c_i; `R` stands for red and `W` stands for white.

You can do the following two kinds of operations any number of times in any order:

* Choose two stones (not necessarily adjacent) and swap them.
* Choose one stone and change its color (from red to white and vice versa).



According to a fortune-teller, a white stone placed to the immediate left of a red stone will bring a disaster. At least how many operations are needed to reach a situation without such a white stone?

Constraints

* 2 \leq N \leq 200000
* c_i is `R` or `W`.

Input

Input is given from Standard Input in the following format:


N
c_{1}c_{2}...c_{N}


Output

Print an integer representing the minimum number of operations needed.

Examples

Input

4
WWRR


Output

2


Input

2
RR


Output

0


Input

8
WRWWRWRR


Output

3
//Created by Aminul on 8/2/2020.

import sun.reflect.generics.tree.Tree;

import java.io.*;
import java.util.*;
import static java.lang.Math.*;

public class Main {
    public static void main(String[] args)throws Exception {
        Scanner in = new Scanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        int n = in.nextInt();
        char[] str = in.next().toCharArray();
        TreeSet<Integer> red = new TreeSet<>(), white = new TreeSet<>();
        for (int i = 0; i < n; i++) {
            if(str[i] == 'R') red.add(i);
            else white.add(i);
        }

        int cnt = 0;
        while (!white.isEmpty() && !red.isEmpty() && white.first() < red.last()) {
            int r = red.pollLast(), w = white.pollFirst();
            red.add(w);
            white.add(r);
            cnt++;
        }

        pw.println(cnt);
          
        pw.close();
    }

    static void debug(Object...obj) {
        System.err.println(Arrays.deepToString(obj));
    }
}