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 <stdlib.h>
#include <math.h>
#include <string.h>
 
 
void swap (int *x, int *y) {
  int temp;    
 
  temp = *x;
  *x = *y;
  *y = temp;
}
 
int gcd(int a,int b){
  if(a<b){
    swap(&a,&b);
  }
 
  if(b==0){
    return a;
  }
 
  int r=a%b;
  while(r!=0){
    a = b;
    b = r;
    r = (a%b);
  }
 
  return b;
}
 
/* クイックソート */
int dn(const void*a,const void*b){return*(int*)b-*(int*)a;}
 
 
int modpow(int a,int n,int mod){
  int ans=1;
  while(n>0){
    if(n & 1){
      ans=ans*a%mod;
    }
    a=a*a%mod;
    n/=2;
  }
  return ans;
}
 
int max(int a,int b){
  if(a<b){
    return b;
  }
  else{
    return a;
  }
}
 
int min(int a,int b){
  if(a<b){
    return a;
  }
  else{
    return b;
  }
}
 
void chmax(int *a,int b){
  if(*a < b){
    *a = b;
  }
 
  return;
}
 
void chmin(int *a,int b){
  if(*a > b){
    *a =b;
  }
 
  return;
}
 

int main(void){
  int n;
  scanf("%d", &n);
  char c[n];
  scanf("%s", c);
  int r=0;
  int w=0;
  int i;
  for(i=0; i<n; i++){
    if(c[i] == 'W'){
      w++;
    }
    if((n/2) <= i && c[i] == 'R'){
      r++;
    }
  }
  if(w==0){
    printf("0\n");
    return 0;
  }else{
    printf("%d\n", r);
    return 0;
  }
}