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: Wrong Answer)

#include<stdio.h>
#include<math.h>
int main (void){
    unsigned long int n;
    unsigned long int cnt = 0;
    char c[200003];
    int allW = 1;
    int allR = 1;
    int plane = 0;
    int renzokuW = 0;
    scanf("%ld\n",&n);
    unsigned long int fixN = n;
    for(int i=0;i<n;i++){
        scanf("%c",&c[i]);
        if(c[i] == 'W'){
            allR = 0;
            if(plane == 1)plane = 2;
        }
        if(c[i] == 'R'){
            fixN = i+1;
            if(i!= 0)allW = 0;
            if(plane == 0)plane = 1;
            if(plane == 2)plane = 3;
        }
    }    
    if(allR == 1 || allW == 1 || plane == 2){
        printf("0");
        return 0;
    }
    unsigned long int ref = (float)fixN/2+0.5;
    for(int i = ref;i<fixN;i++)if(c[i] == 'R')cnt ++;

    printf("%ld",cnt);
    return 0;
}