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>
using namespace std;

using ll = long long;
using P = pair<ll, ll>;

const ll MOD = 1e9+7;
// const ll MOD = 998244353;
const ll INF = 1ll<<60;

#define FOR(i,a,b) for (ll i=(a);i<(ll)(b);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
#define DEBUG(x) ;
#define DEBUG(x) std::cerr << #x << " : " << (x) << std::endl;

int dx[4]{0, 1, 0, -1};
int dy[4]{1, 0, -1, 0};



int main(int argc, char **argv)
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	ll N; cin >> N;
	string s; cin >> s;
	ll res{0};
	ll res2{0};
	bool ok{false};
	for (auto c : s)
	{
		if (c == 'W')
			ok = true;

		res2 += c == 'W';
		if (ok)
		{
			res += c == 'R';
		}
	}

	string tmp = s.substr(res);
	ll cnt{0};
	for (auto c : tmp)
	{
		cnt += (c=='W');
	}

	std::cout << min(max(0ll, res2-cnt), res) << std::endl;
	return 0;
}