Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

4
WWRR

Sample Output 1

2

For example, the two operations below will achieve the objective.

Swap the 1-st and 3-rd stones from the left, resulting in RWWR.

Change the color of the 4-th stone from the left, resulting in RWWW.

Sample Input 2

2
RR

Sample Output 2

0

It can be the case that no operation is needed.

Sample Input 3

8
WRWWRWRR

Sample Output 3

3

Solution in Java (OpenJDK 11.0.6) (judge verdict: Runtime Error)

import java.util.*;

public class Main {

    static int gcd (int a, int b) {return b>0?gcd(b,a%b):a;}
    static long mod = 1000000007;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        char[] c = sc.next().toCharArray();
        boolean hasWhite = false;
        int ans = 0;
        int redCounter= N-1;

        for(int i=0;i<N;i++){
            if(c[i] == 'W'){
                boolean isFind = false;
                while(!isFind){
                    if(c[redCounter] == 'R'){
                        if(i < redCounter){
                            ans++;
                        }
                        isFind = true;
                    }
                    redCounter--;
                }

            }
        }

        System.out.println(ans);


    }

}