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)

// Code for D - Alter Altar
#include <bits/stdc++.h>

using namespace std;

typedef long long ll;

#define REP(i,n) for(ll i=0;i<(ll)(n);i++)
#define FOR(i,x,y) for(ll i=x;i<(ll)(y);i++)

#define debug(var)  do{std::cout << #var << " : ";view(var);}while(0)
template<typename T> void view(T e){std::cout << e << std::endl;}
template<typename T> void view(const std::vector<T>& v){for(const auto& e : v){ std::cout << e << " "; } std::cout << std::endl;}
template<typename T> void view(const std::vector<std::vector<T> >& vv){ for(const auto& v : vv){ view(v); } }

int main() {
  int N;
  string c;
  cin >> N;
  cin >> c;

  int rc = 0;
  int wc = 0;

  REP(i, N) {
    if (c[i] == 'R') {
      rc++;
    } else {
      wc++;
    }
  }

  int rcd = 0;
  int wcd = 0;
  REP(i, rc) {
    if (c[i] == 'W') {
      rcd++;
    }
  }

  REP(i, wcd) {
    if (c[rc + i] == 'R') {
      wcd++;
    }
  }

  cout << max(rcd, wcd) << endl;
  return 0;
}