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)

#pragma GCC optimize("Ofast")
#define _USE_MATH_DEFINES
#include "bits/stdc++.h"

using namespace std;

using u8 = uint8_t;
using u16 = uint16_t;
using u32 = uint32_t;
using u64 = uint64_t;
using i8 = int8_t;
using i16 = int16_t;
using i32 = int32_t;
using i64 = int64_t;

constexpr char newl = '\n';
constexpr double eps = 1e-10;

#define FOR(i,a,b) for (int i = (a); i < (b); i++)
#define F0R(i,b) FOR(i,0,b)
#define RFO(i,a,b) for (int i = ((b)-1); i >=(a); i--)
#define RF0(i,b) RFO(i,0,b)
#define fi first
#define se second
#define debug(x) cout << #x << ": " << x << '\n';
#define rng(a) a.begin(),a.end()
#define rrng(a) a.rbegin(),a.rend()

template<typename T1, typename T2> inline void chmin(T1& a, T2 b) { if (a > b) a = b; }
template<typename T1, typename T2> inline void chmax(T1& a, T2 b) { if (a < b) a = b; }
template<class T> void Print(vector<T> v) {
  F0R(i, v.size()) {
    cout << v[i] << ' ';
  }
  cout << newl;
}

#if 1


// INSERT ABOVE HERE
signed main() {
  cin.tie(0);
  ios_base::sync_with_stdio(false);

  int N;
  string S;
  cin >> N >> S;
  int l = 0, r = S.size() - 1;
  int result = 0;
  while (true) {
    while (l < S.size() && S[l] == 'R') l++;
    while (r >= 0 && S[r] == 'W') r--;
    if (l <= r) {
      result++;
      l++;
      r--;
    }
    else {
      break;
    }
  }
  cout << result;
}
#endif