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)

fn main() {
    let n: i64;
    let mut c: Vec<u8>;
    let mut buf: String;
    buf = String::new();
    let stdin_obj = std::io::stdin();
    stdin_obj.read_line(&mut buf).ok();
    n = buf.trim().parse().ok().unwrap();
    stdin_obj.read_line(&mut buf).ok();
    c = Vec::<u8>::with_capacity(n as usize);
    for item in buf.trim().as_bytes() {
        c.push(*item);
    }
    let mut counter: i32 = 0;
    for i in 0..c.len() {
        if c[i] == 'R' as u8 {
            continue;
        }
        let mut has_remaining: i32 = -1;
        for j in i + 1..c.len() {
            if c[j] == 'W' as u8 {
                if has_remaining < 0 {
                    c[j] = 'R' as u8;
                    counter += 1;
                }
                has_remaining += 1;
            }
        }
        if has_remaining <= 0 {
            break;
        }
    }
    println!("{}", counter);
}