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>
#include <assert.h>
#include <stdlib.h>
#include <string.h>

int main(void)
{
	int n;//文字数
	char dummy[100];
	char *c;//石
	char tmp;
	int ac=0;//答のカウント
	int rfc=0;//最初のRの位置
	int i,j;
	scanf("%d",&n);
	c=(char *)malloc(sizeof(char)*(n+1));
	scanf("%c",&dummy);//point
	for(i=0;i<n;i++)
	{
		scanf("%c",&c[i]);
	}
	for(i=0;i<n;i++)//最初のRの位置
	{
		if(c[i]=='R')
		{
			rfc=i;
			break;
		}
	}
	//printf("%d\n",rfc);
	for(i=rfc;i<n;i++)
	{
		if(rfc==ac)
		{
			break;
		}
		if(c[i]=='R')
		{
			tmp=c[ac];
			c[ac]=c[i];
			c[i]=tmp;
			ac++;
		}
	}
	for(i=rfc-1;i<n-1;i++)
	{
		if(c[i]=='W')
		{
			if(c[i+1]=='R')
			{
				c[i+1]=='W';
				ac++;
			}
		}
	}
	/*
	for(i=0;i<n;i++)//操作後の配列
	{
		printf("%c",c[i]);
	}
	*/
	//printf("\n");
	printf("%d\n",ac);//操作回数
	free(c);
	return 0;
}