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 <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<n; i++)
#define REP(i,m,n) for(ll i=(ll)(m);i<(ll)(n);i++)
#define fi first
#define se second
long long mo = 1e9 + 7;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> Pii;
typedef pair<ll,ll> Pll;
template<class T, class S> void cmin(T &a, const S &b) { if (a > b)a = b; }
template<class T, class S> void cmax(T &a, const S &b) { if (a < b)a = b; }
template<class A>void PR(A a,ll n){rep(i,n){if(i)cout<<' ';cout<<a[i];}cout << "\n";}
ld PI=3.14159265358979323846;

int main(){
    ll N;
    string S;
    cin >> N >> S;
    //ll l = 0;
    ll r = N-1;
    ll ans = 0;
    rep(l,N){
        if(S[l] == 'W'){
            while(S[r] != 'R' && l < r){
                r--;
            }
            if(!(l<r)){
                break;
            }
            S[l] = 'R';
            S[r] = 'W';
            ans++;
            r--;
        }
        if(!(l<r))break;
    }
    cout << ans << endl;
}