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>
using namespace std;

// #define int long long // intで書いたけど心配なときにlong longに変換する
struct Fast {Fast(){std::cin.tie(0);ios::sync_with_stdio(false);}} fast;

/* short */
#define pb push_back
#define eb emplace_back
#define mp make_pair
#define Fi first
#define Se second
#define ALL(v) begin(v), end(v)
#define RALL(v) rbegin(v), rend(v)
#define X real()
#define Y imag()

/* REPmacro */
#define REPS(i, a, n) for (ll i = (a); i < (ll)(n); ++i)
#define REP(i, n) REPS(i, 0, n)
#define RREP(i, n) REPS(i, 1, n + 1)
#define DEPS(i, a, n) for (ll i = (a); i >= (ll)(n); --i)
#define DEP(i, n) DEPS(i, n, 0)
#define EACH(i, n) for (auto&& i : n)

/* debug */
#define debug(x) cerr << x << " " << "(L:" << __LINE__ << ")" << '\n';

/* alias */
using ll = long long;
using ull = unsigned long long;
using uint = unsigned int;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using pii = pair<int, int>;
using tiii = tuple<int, int, int>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using plglg = pair<ll, ll>;
using tlglglg = tuple<ll, ll, ll>;
using D = double;
using P = complex<D>;
using vs = vector<string>;
template <typename T> using PQ = priority_queue<T>;
template <typename T> using minPQ = priority_queue<T, vector<T>, greater<T>>;

/* const */
const int INF = 1001001001;
const ll LINF = 1001001001001001001ll;
const int MOD = 1e9 + 7;
const D EPS = 1e-9;
const int dx[] = {0, 1, 0, -1, 1, -1, 1, -1}, dy[] = {1, 0, -1, 0, 1, -1, -1, 1};

/* func */
inline bool inside(int y, int x, int H, int W) {return y >= 0 && x >= 0 && y < H && x < W;}
inline int in() {int x; cin >> x; return x;}
inline ll IN() {ll x; cin >> x; return x;}
inline vs split(const string& t, char c) {vs v; stringstream s(t); string b; while(getline(s, b, c)) v.eb(b); return v;}
template <typename T> inline bool chmin(T& a, const T& b) {if (a > b) a = b; return a > b;}
template <typename T> inline bool chmax(T& a, const T& b) {if (a < b) a = b; return a < b;}
template <typename T, typename S> inline void print(const pair<T, S>& p) {cout << p.first << " " << p.second << endl;}
template <typename T> inline void print(const T& x) {cout << x << endl;}
template <typename T, typename S> inline void print(const vector<pair<T, S>>& v) {for (auto&& p : v) print(p);}
template <typename T> inline void print(const vector<T>& v, string s = " ") {REP(i, v.size()) cout << v[i] << (i != (ll)v.size() - 1 ? s : "\n");}
inline void yesno(bool ok, string yes = "Yes", string no = "No") {cout << (ok?yes:no) << endl;}


signed main() {
    //my-code
    ll n; cin >> n;
    string s; cin >> s;

    ll ans = 0;
    ll l = 0, r = n-1;
    while (l<r) {
        if (s[l] != 'W') {
            l++;
            continue;
        }
        if (s[r] != 'R') {
            r--;
            continue;
        }
        if (s[l] == 'W' && s[r] == 'R') {
            ans++;
            l++;
            r--;
        }
    }

    print(ans);
    return 0;
}

// https://github.com/kurokoji/.cpp-Template/wiki テンプレートについて
// http://www.creativ.xyz/dump-cpp-652 dump()について
// https://gist.github.com/rigibun/7905920 色々