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: Accepted)

use proconio::marker::Chars;
use proconio::{fastout, input};
use std::{cmp::Reverse, collections::BinaryHeap};

// 最終的に R..RW..W という状態にする
// WRWWRWRR
// RRWWRWRW
// RRRWRWWW
// RRRRWWWW

#[fastout]
fn main() {
    input! {
        n: usize,
        c: Chars,
    }

    // rは添字の大きい順に並べる
    let mut r = BinaryHeap::new();
    // wは添字の小さい順に並べる
    let mut w = BinaryHeap::new();
    for i in 0..n {
        if c[i] == 'R' {
            r.push(i);
        } else {
            w.push(Reverse(i));
        }
    }

    let mut count = 0i64;
    loop {
        let r_rightmost = r.pop();
        let w_leftmost = w.pop();
        match (r_rightmost, w_leftmost) {
            (None, _) | (_, None) => {
                break;
            },
            (Some(r_rightmost), Some(Reverse(w_leftmost))) => {
                if r_rightmost < w_leftmost {
                    break;
                }
            }
        }
        count += 1;
    }

    println!("{}", count);
}