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 fastIO ios::sync_with_stdio(0), cin.tie(0)
#define endl '\n'
#define ft first
#define sd second
#define pb push_back
#define pob pop_back()
#define pf push_front
#define pof pop_front()
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin, (x).rend()
typedef long long ll;
typedef pair<int,int> pii;
typedef pair<char,char> pcc;
typedef pair<double,double> pdd;
typedef pair<long long,long long> pll;
// scanf("%d %d %d %d",&w,&b,&d,&s); %lld
// printf("%d\n" ,points);
// const double PI=acos(-1);

int main(){
    int n,r=0,rta=0;
    string s;
    cin >> n;
    cin >> s;
    for(int i=0;i<n;i++){
        if(s[i]=='R') r++;
    }
    for(int i=0;i<r;i++){
        if(s[i]!='R') rta++;
    }
    cout << rta << endl;
    return 0;
}