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>
#define rep(i,N) for(int i=0;i<(N);i++)
#define rrep(i, n) for (int i = (int)n-1; i >= 0; --i)
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
using namespace std;
const long long MOD = 1e9 + 7;
const long long INF = 1e12;
const int inf = 1e9;
const int mod = 1e9+7;
typedef long long ll;
typedef pair<ll,int> P;
typedef set<int> S;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }

__attribute__ ((constructor))
void init() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout << fixed << setprecision(10);
}

signed main(){
  int n;
  string s;
  cin >> n >> s;
  int cnt_r = 0;
  rep(i,int(s.size())){
    if(s[i] == 'R')
      cnt_r++;
  }
  int ans = 0;
  rep(i,cnt_r){
    if(s[i] == 'W')
      ans++;
  }
  cout << ans << endl;
  return 0;
}