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)

#include <bits/stdc++.h>
using namespace std;

#ifdef LOCAL
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...) 42
#endif
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
    cerr << name << ": " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
    const char* comma = strchr(names + 1, ',');
    cerr.write(names, comma - names) << ": " << arg1 << "\t|";
    __f(comma + 1, args...);
}

#define print(x) for(auto it : x) cout<< it << ' '; cout << '\n';
typedef long long ll;
typedef unsigned long long ull;
#define mem1(a)           memset(a,-1,sizeof(a))
#define mem0(a)           memset(a,0,sizeof(a))

int main() {
    ios::sync_with_stdio(false); cin.tie(0);
    int n; cin >> n;
    string s; cin >> s;
    int i = 0, j = n - 1;
    int ans = 0;
    while (i < j) {
        if (s[i] == 'W') {
            while (j > i && s[j] != 'R') {
                j--;
            }
            swap(s[i], s[j]);
            ans++;
        }
        i++;
    }
    cout << ans << '\n';
    return 0;
}