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 Rust (1.42.0) (judge verdict: Time Limit Exceeded)

use proconio::input;
use proconio::marker::{Chars};

// x ** n % mods を求める
pub fn mod_pow(x: i64, n: i64, mods: i64) -> i64 {
    if n == 0 {
        return 1;
    }
    let mut res = mod_pow(x * x % mods, n / 2, mods);
    if n % 2 == 1 {
        res = res * x % mods;
    }
    return res;
}

// (1/n) % mods を求める
pub fn mod_inv(n: i64, mods: i64) -> i64 {
    return mod_pow(n, mods-2, mods);
}

// nCk % mods を求める
pub fn mod_comb(n: i64, k: i64, mods: i64) -> i64 {
    let sum1 = (n-k+1..n+1).fold(1, |ac, x| ac * x % mods);
    let sum2 = (1..k+1).fold(1, |ac, x| ac * x % mods);
    return sum1 * mod_inv(sum2, mods) % mods;
}

// nCk の組み合わせを列挙する
pub fn combination(n: usize, k: usize) -> Vec<Vec<usize>> {
    let mut ans = Vec::new();
    if k == 1 {
        for i in 1..n+1 {
            let mut tmp = Vec::new();
            tmp.push(i);
            ans.push(tmp);
        }
    }
    else {
        for iter in combination(n-1, k-1) {
            let max = iter.iter().max().unwrap();
            for j in *max+1..n+1 {
                let mut tmp = iter.clone();
                tmp.push(j);
                ans.push(tmp);
            }
        }
    }
    ans
}

use std::collections::VecDeque;

fn main() {
    input!{
        n: usize,
        mut c: Chars,
    }
    let mut ans = 0;
    let mut bufr = VecDeque::new();
    let mut bufw = VecDeque::new();
    for i in (0..n).rev() {
        if c[i] == 'R' {
            bufr.push_back(i);
        }
    }
    for i in 0..n {
        if c[i] == 'W' {
            bufw.push_back(i);
        }
    }
    loop {
        let mut flag = 0;
        if bufr.len() > 0 {
            for i in (1..bufr[0]+1).rev() {
                if c[i] == 'R' && c[i-1] == 'W' {
                    c.swap(bufw.pop_front().unwrap(), bufr.pop_front().unwrap());
                    flag = 1;
                    ans += 1;
                    break;
                }
            }
        }
        if flag == 0 {
            println!("{}", ans);
            return;
        }
    }
}