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 fi first
#define se second
#define rep(i, n) for(int (i) = 0; (i) < (n); (i)++)
#define rrep(i, n) for (int (i) = 1; (i) <= (n); (i++))
#define rng(a) a.begin(),a.end()
#define rrng(a) a.rbegin(),a.rend()
#define pb push_back
#define eb emplace_back
#define yn {puts("Yes");}else {puts("No");}
#define printd(x) printf("%.20f\n", x);
template<class T> bool chmin(T& a, const T& b) {if (a > b) {a = b; return 1;} return 0;}
template<class T> bool chmax(T& a, const T& b) {if (a < b) {a = b; return 1;} return 0;}
using namespace std;
using ll = long long;

int main() {
	int n;
	cin >> n;
	string s;
	cin >> s;

	int cntw = 0;
	rep (i, n) if (s[i] == 'W') cntw++;

	int ans = 0;
	rep (i, cntw) if (s[i] == 'W') ans++; 

	cout << ans << endl;
	return 0;
}