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: Output Limit Exceeded)

#include<stdio.h>
int N;
char stone[200000];
int Rn[200000];
int R=0;
int count=0;
void swap(int a, int b){
  if(a==b)return;
  char w = stone[a];
  stone[a]=stone[b];
  stone[b]=w;
  count++;
}
int main(void){
  int c;
  int i;
  int j;
  int k;
  scanf("%d",&N);
  i=0;j=0;
  c=getchar();
  while(c!=EOF){
    if( c=='R' || c=='W'){
      stone[i]=c;
      if(c=='R'){
        R++;
        Rn[j++]=i;
        printf("Rn[%d]の場所は%d\n",j-1,Rn[j-1]);
      }
      i++;
    }
    c=getchar();
  }
  stone[i]='\0';
  k=j; j=0;
  for(i=0;i<R;i++){
    if(stone[i]=='R')
      j++;
   }
  if(j==k){
    printf("%d\n",count);
    return 0;
  }
  for(i=0;i<N;i++){
    if(stone[i]='W'){
      swap(i,Rn[j++]);
      printf("stone=%s\n",stone);
    }
    if(j==k)break;
  }
  printf("%d\n",count);
  return 0;
}