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::collections::BinaryHeap;
use std::cmp::Reverse;

fn main() {
    let n: usize = {
        let mut buf = String::new();
        std::io::stdin().read_line(&mut buf).unwrap();
        buf.trim_end().parse().unwrap()
    };

    let c: Vec<char> = {
        let mut buf = String::new();
        std::io::stdin().read_line(&mut buf).unwrap();
        buf.trim_end().chars().collect()
    };

    let mut rheap = BinaryHeap::new();
    let mut wheap = BinaryHeap::new();
    for i in 0..n {
        if c[i] == 'R' {
            rheap.push(i);
        } else {
            wheap.push(Reverse(i));
        }
    }

    let mut ans = 0;
    while !rheap.is_empty() && !wheap.is_empty() {
        let x = rheap.peek().unwrap();
        let Reverse(y) = wheap.peek().unwrap();
        if x < y {
            break;
        } else {
            ans += 1;
            rheap.pop();
            wheap.pop();
        }
    }
    println!("{}", ans);
}