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>
#include <variant>
#include <boost/range/algorithm.hpp>
#include <boost/range/numeric.hpp>
#include <boost/range/irange.hpp>
#include <boost/range/adaptor/indexed.hpp>
#define FOR(i,k,n)  for(int i = (k);i < (n);++i)
#define REP(i,n)    FOR(i,0,n)
#define ALL(x) begin(x),end(x)

using namespace std;
using namespace boost::adaptors;
using namespace std::string_literals;
using ll = int64_t;
using vecint = vector<int>;
using vecll = vector<ll>;
using boost::irange;

int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin>>n;
  string s;
  cin>>s;
  int cntr = 0;
  for(auto&&c:s) {
    if (c == 'R') ++cntr;
  }
  int cntw = 0;
  int mn = cntr;
  for(auto&&c:s) {
    if (c == 'R') --cntr;
    else ++cntw;
    mn = min(mn, max(cntr, cntw));
  }
  cout<<mn<<endl;
  return 0;
}