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 on: 2 / 8 / 2020
 *  Author:     Eslam Waheed
 */

#define _CRT_SECURE_NO_WARNINGS
//#define __USE_MINGW_ANSI_STDIO

#include <bits/stdc++.h>
//#include <bits/BIGINT.h>
//#include <bits/Int128.h>
using namespace std;

#define Eslam__Waheed ios_base::sync_with_stdio(false);cout.tie(NULL);cin.tie(NULL);

#define ll   long long
#define ull  unsigned ll
#define dd   double
#define ld   long dd
#define ss   string

#define pb      push_back
#define all(c)  c.begin(), c.end()
#define rall(c) c.rbegin(), c.rend()
#define stp(n)  cout << fixed << setprecision(n);
#define iq(v)   v.resize(unique(v.begin(),v.end())-v.begin())
#define sz(s)   (ll)(s.size())
#define endl    '\n'

#define loop(i,m,n)  for(ll i=m;i<n;i++)
#define test(t)      while(t--)

//ll gcd(ll a, ll b) { return !b ? abs(a) : gcd(b, a % b); }
//ll lcm(ll a, ll b) { return abs(a / gcd(a, b)) * b; }

//const ld PI = acos(-1);
//const ll mod = 1e9 + 7;
//const ld EPS = 1e-8;
/*
 you can ^_^ (•‿•)

( إن الله وملائكته يصلون على النبي يا أيها الذين آمنوا صلوا عليه وسلموا تسليما )
*/
int main()
{
	//stp(0)
	Eslam__Waheed;
	ll x; cin >> x;
	ss s; cin >> s;
	ll r = count(all(s), 'R'), a = 0;
	loop(i, 0, r)
	{
		if (s[i] == 'R')a++;
	}
	if (a > 0)
		cout << r - a;
	else
		cout << r;
}