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<queue>
#include<memory>
#include<map>
#include<set>
#include<unordered_map>
#include<array>
#include<vector>
#include<algorithm>
#include<list>
#include<limits>
#include<stdio.h>
#include<math.h>
#include<string>
#include<fstream>
#include<functional>


using namespace std;
using ll = long long;
using vec = vector<int>;
using mat = vector<vec>;
using lst = list<int>;
using pr  = pair<int, int>;


int main() 
{
	int N;
	string c;
	cin >> N >> c;

	int l = -1, r = N;
	int i = 0;
	for (i = 0;l<r; ++i)
	{
		
		do
		{ 
			l++; 
			if(l==N)
			{
				cout << i << endl; return 0;
			}
		}while (c[l] != 'W');

		do
		{
			r--;
			if (r == -1)
			{
				cout << i << endl; return 0;
			}
		} while (c[r] != 'R');

		c[l] = 'R';
		c[r] = 'W';
	}

	cout << i-1 << endl;
	return  0;

}