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)

#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
 
#include <bits/stdc++.h>
using namespace std;
 
typedef long long ll;
typedef long double ld;
 
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;
 
typedef vector<int> vi;
typedef vector<ld> vd;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
 
#define IOS ios_base::sync_with_stdio(0); cin.tie(0)
#define FINOUT freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout)
#define FOR(i, a, b) for (int i = a; i < (b); ++i)
#define FORd(i, a, b) for(int i = (b) - 1; i >= a; --i)
#define trav(a, x) for(auto& a : x)

#define sz(x) (int)(x).size()
#define mp make_pair
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define all(x) x.begin(), x.end() 
#define endl '\n'
const int MOD = 1000000007;

int main(){
    IOS;
    // FINOUT;
    int T = 1;
    int n;
    cin >> n;
    string s;
    cin >> s;
    int cnt = 0;
    FOR(i, 0, n / 2) if (s[i] == 'W') cnt++;
    cout << cnt;
    return 0;
}