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 std::io;
use std::collections::VecDeque;
use std::process::exit;

fn main() {
    let mut s = String::new();
    io::stdin().read_line(&mut s).ok();
    let n: usize = s.trim().parse().ok().unwrap();
    let mut s = String::new();
    io::stdin().read_line(&mut s).ok();

    let (mut r, mut w) = (VecDeque::new(), VecDeque::new());
    r.push_back(0);
    for (i, c) in s.trim().chars().enumerate() {
        match c {
            'R' => r.push_back(i),
            'W' => w.push_front(i),
            _ => exit(1),
        };
    }
    w.push_front(n);
    let c = r.iter().rev().zip(w.iter().rev())
        .filter(|(x, y)| x > y)
        .count();
    println!("{}", c);
}