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 C++ (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <iostream>
#include <string>

using namespace std;

string exchange(string c, int i, int j){
    string temp;
    temp = c[i];
    c[i] = c[j];
    c[j] = temp[0];

    return c;
}

string change(string c, int i){

    if(c[i] == 'W'){c[i] = 'R';}
    else{c[i] = 'W';}

    return c;
}

int judge(string c){
    int t = 0;
    int length = c.length();
    int has_R = 0;
    
    for(int i = 0; i < length; i++){ //count a number of 'R'
        if(c[i] == 'R'){
            has_R++;
        }
    }


    for (int i = has_R; i < length; i++){
        if(c[i] == 'R'){
            for(int j = 0; j < has_R; j++){
                if(c[j] == 'W'){
                    c = exchange(c, j, i);
                    t++;

                    break;
                }
            }
        }
        has_R = i;
    }


    return t;
}



int main(void){
    int N;
    string c;

    cin >> N;
    cin >> c;

    int ans;

    ans = judge(c);
    cout << ans;
    return 0;
}