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<bits/stdc++.h>
#define ull unsigned long long
# define ll long long
# define mod 1000000007LL 
using namespace std;
int main()
{
     ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll n;
    cin>>n;
    string s;
    cin>>s;
    stack<char>ans;
    ll count=0;
    for(int i=0;i<n;i++)
    {
        if(s[i]=='W' && ans.size()==0)
        ans.push('W');
        else if(s[i]=='R' && ans.size()!=0)
        {
            ans.pop();
            count++;
        }

    }
    for(int i=0;i<n-3;i++)
    {
        if(s[i]=='W' && s[i+1]=='W' && s[i+2]=='R' && s[i+3]=='R')
        count++;
    }
    
    cout<<count;



    
    return 0;
}