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 ll = long long;
#define FOR(i, k, n) for(ll i = (k); i < (n); i++)
#define FORe(i, k, n) for(ll i = (k); i <= (n); i++)
#define FORr(i, k, n) for(ll i = (k)-1; i > (n); i--)
#define FORre(i, k, n) for(ll i = (k)-1; i >= (n); i--)
#define REP(i, n) FOR(i, 0, n)
#define REPr(i, n) FORre(i, n, 0)
#define ALL(x) (x).begin(), (x).end()
#define ALLr(x) (x).rbegin(), (x).rend()
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
using namespace std;

const int INF = 1001001001;

int main(void){
  ll n;
  cin >> n;
  string s;
  cin >> s;
  ll a = 0, b = 0;
  REP(i, n){
    if(s[i] == 'R') a++;
  }
  ll ans = max(a, b);
  REP(i, n){
    if(s[i] == 'R') --a;
    else b++;
    ll now = max(a, b);
    chmin(ans, now);
  }
  cout << ans << endl;
  return 0;
}