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: Wrong Answer)

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

typedef long long ll;
typedef vector<ll> vll;
typedef vector<int> vz;
typedef vector<vz> vvz;
typedef pair<ll, ll> pll;
#define sz(x) int((x).size())
#define fo(i,n) for(ll i=0; i<(n); i++)
#define FO(i,a,b) for(ll i=(a); i!=(b); i+=(b)>(a)?1:-1)
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define all(c) c.begin(), c.end()
#define F first
#define S second
#define os(x) cout<<x<<' '
#define on(x) cout<<x<<"\n"
#define nl cout<<"\n"
#define maxe(x) max_element(all(x))-(x).begin()
#define mine(x) min_element(all(x))-(x).begin()

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL);
#ifndef ONLINE_JUDGE
	freopen("C:\\Users\\ASISH\\Desktop\\coding\\input1.txt", "r", stdin);
	freopen("C:\\Users\\ASISH\\Desktop\\coding\\output1.txt", "w", stdout);
#endif

	int tc=1;
	// cin>>tc;
	fo(q, tc)
	{
		ll n, cnt=0;
      	string s;
      	cin>>n>>s;
      	fo(i, n-1){
          if(s[i]=='W' && s[i+1]=='R') s[i+1]='W', cnt++;
        }
      	on(cnt);
	}
	cerr << "Time : " << ((double)clock()) / (double)CLOCKS_PER_SEC << "s\n";
	return 0;
}