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: Accepted)

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

#define RED 'R'
#define WHITE 'W'

char stone_int_array[200000];

int main()
{
    int n, exchange_num = 0;
    char not_stone;
    char scan_stone;
    int left_counter, right_counter;

    scanf("%d", &n);

    scanf("%c", &not_stone);

    for (int i = 0; i<n; i++)
    {
        scanf("%c", &stone_int_array[i]);

    }

    left_counter = 0;
    right_counter = n-1;

    while (left_counter < right_counter)
    {
        for (; stone_int_array[left_counter] == RED; left_counter++)
        { }

        for (; stone_int_array[right_counter] == WHITE; right_counter--)
        { }

        if (left_counter < right_counter)
        {
            exchange_num ++;
            left_counter ++;
            right_counter --;
        }
    }
    


    printf("%d", exchange_num);

    return 0;
}