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 <bits/stdc++.h>
#include <stdlib.h>
#include <fstream>
#include <iostream>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
typedef pair<ll,ll> llP;

ll mod(ll a, ll b){
  ll ret=a%b;
  if(ret<0) ret+=b;
  return ret;
}

ll modpow(ll a,ll b,ll c){
  ll res=1;
  while(b>0){
    if(b&1) res=mod(res*a,c);
    a=mod(a*a,c);
    b>>=1;
  }
  return res;
}

int main() {
  int n;
  cin>>n;
  char c[n];
  int sum[n];
  for(int i=0;i<n;i++){
    cin>>c[i];
    if(i==0){
      if(c[i]=='R') sum[i]=1;
      else sum[i]=0;
    }
    else{
      if(c[i]=='R') sum[i]=sum[i-1]+1;
      else sum[i]=sum[i-1];
    }
  }
  if(sum[n-1]==0 or sum[n-1]==n){
    cout<<0<<endl;
    return 0;
  }
  cout<<sum[n-1]-sum[sum[n-1]-1]<<endl;
  return 0;
}