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 (Clang 10.0.0) (judge verdict: Time Limit Exceeded)

#include<stdio.h>
#include<string.h>

int main(void){

    int N;
    int i,j;
    int once,second;
    char temp;
    int count;
    int sum = 0;
    int a = 0;

    scanf("%d",&N);
    char c[N];

    for(i = 0;i < N;i++){
        scanf(" %c",&c[i]);
    }

    count = N;
    while(count--){
        for(i = 0;i < N;i++){

            a = 0;

            if(c[i] == 'W'){

                once = i;

                for(j = i + 1;j < N;j++){

                    if(c[j] == 'R'){

                        second = j;
                        a++;
                    }
                    if(j == N - 1){

                        if(a == 1){
                            c[second] = 'W';
                            sum++;
                            break;

                        }else if(a > 1){
                            temp = c[second];
                            c[second] = c[once];
                            c[once] = temp;

                            sum++;
                            break;
                        }
                    }
                }
            }
        }
    }

    printf("%d\n",sum);

    return 0;
}