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: Runtime Error)

#include <bits/stdc++.h>
#include <math.h>
using namespace std;
typedef long long ll;
typedef long double ld;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
const ll mod = 1000000007;
const ll INF = 1001001001;
const ll LINF = 1001001001001001001;

void prvec(vector<ll> vec){
    ll n = vec.size();
    rep(i,n) cout << i << " " << vec.at(i) << "\n";
}

void pr2d(vector<vector<ll>> vvec){
    ll h = vvec.size();
    ll w = vvec.at(0).size();
    rep(i,h){
        rep(j,w){
            cout << vvec.at(i).at(j) << " ";
        }
        cout << "\n";
    }
}

int main(){
    ll n ; cin >> n;
    string s; cin >> s;
    ll l = 0, r = n-1, cnt = 0;
    while(r-l>1){
        while(s.at(l)=='R'){
            l++;
        }
        s.at(l) = 'R';
        while(s.at(r) == 'W'){
            r--;
        }
        s.at(r) = 'W';
        cnt++;
        //cout << s << endl;
    }
    cout << cnt << endl;
    return 0;
}