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)

/*
 *  Created by Andrei Arhire 02/08/2020
 *  Copyright © 2020 Andrei Arhire. All rights reserved.
 */
#include <bits/stdc++.h>
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define lll long long
#define llf __float128
#define lld long double
using namespace std;
const lll NR = 1e5 + 10  ,oo = 2e9 + 10, MOD = 1e9 + 7 ;
const long double pi = 2*acos(0.0);

string s ;
int red , white , ans, n  ;

signed main () {
    int i ;

    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0) ;
   cin >> n ;
   cin >> s ;
   for ( i = 0 ; i < s.size() ; ++ i )  {
       if ( s [ i ] == 'R' )    red ++ ;
       else                     white ++ ;
   }
   for ( i = 0 ; i < red ; ++ i )   {
       if ( s [ i ] != 'R' )    {
           ans ++ ;
       }
   }
   cout << ans << '\n' ;
    return 0 ;
}