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)

#[allow(unused_imports)]
use proconio::{fastout, input};

#[fastout]
fn main() {
    input!(n: usize, c: String);
    let mut s: Vec<char> = c.chars().collect();

    let mut count = 0;
    let mut index_front = 0;
    let mut index_back = c.len() - 1;
    loop {
        loop {
            if index_front > n - 1 || s[index_front] == 'W' {
                break;
            }
            index_front += 1;
        }
        loop {
            if index_back == 0 || s[index_back] == 'R' {
                break;
            }
            index_back -= 1;
        }

        if index_front > n - 1 || index_back == 0 || index_front >= index_back {
            break;
        }

        //交換
        // indexが交差していた場合終了
        s[index_front] = 'R';
        s[index_back] = 'W';
        count += 1;
    }
    println!("{}", count);
}