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 IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define FRE freopen("input.txt","r",stdin);freopen("output.txt","w",stdout);
#define all(v) (v).begin(),(v).end()
#define endl "\n"
#define int long long
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=a;i>b;i--)
#define vi vector<int>
#define pii pair<int,int>
#define mii map<int,int>
#define SET(A,VAL) memset(A, VAL, sizeof(A))
#define ff first
#define ss second
#define pb push_back
#define eb emplace_back
#define trace1(x)                cerr<<#x<<": "<<x<<endl
#define trace2(x, y)             cerr<<#x<<": "<<x<<" | "<<#y<<": "<<y<<endl
#define trace3(x, y, z)          cerr<<#x<<":" <<x<<" | "<<#y<<": "<<y<<" | "<<#z<<": "<<z<<endl
#define trace4(a, b, c, d)       cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<endl
#define trace5(a, b, c, d, e)    cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<endl
#define trace6(a, b, c, d, e, f) cerr<<#a<<": "<<a<<" | "<<#b<<": "<<b<<" | "<<#c<<": "<<c<<" | "<<#d<<": "<<d<<" | "<<#e<< ": "<<e<<" | "<<#f<<": "<<f<<endl
const int MAX = 1e6 + 8;
const int MOD = 1e9 + 7;
const int sz = 1e5 + 5;
void f_io(){
    #ifndef ONLINE_JUDGE
    FRE
    #endif
}


int32_t main(){
    // f_io();
    IOS;
    int tc = 1;
    //cin>>tc;
    while(tc--){
        int n; cin >> n;
        string s; cin >> s;
        int w = 0, r = 0;
        for(auto k : s) {
            if (k == 'W') w++;
            else r++;
        }
        int sc = 0;
        for(int i = n - 1; i >= n - w; --i) {
            if(s[i] == 'R') sc++;
        }
        // cout << sc << " " << w << " " << r << endl; 
        cout << min(sc, min(w, r));

    }
}