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<iostream>
#include<iomanip>
#include<string>
#include<algorithm>
#include<vector>
#include<utility>
#include<tuple>
#include<map>
#include<queue>
#include<stack>
#include<deque>
#include<bitset>
#include<math.h>
using namespace std;
int main(){
    int n;
    string s;
    cin >> n >> s;

    int w=0,r=0;
    for(int i=0;i<n;i++){
        if(s.at(i)=='W') {
            w++;
        }else{
            r++;
        }
    }
    
    int wr=0;
    if(w<=r){
        for(int i=0;i<w;i++){
            if(s.at(n-1-i)=='W') wr++;
        }
        cout << w-wr << endl;
    }else{
        for(int i=0;i<r;i++){
            if(s.at(i)=='R') wr++;
        }
        cout << r-wr << endl;
    }
}