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: Wrong Answer)

use proconio::input;

#[allow(unused_variables)]
fn main() {
    input! {
      n: usize,
      c: String
    };

    // 先頭以降WRがない場合0
    if !c[1..n].contains("WR") {
        println!("{}", 0);
        return;
    }

    let mut ans = 0;
    let mut flag = false;

    // 先頭WかつRが続く場合、Rが終わるまでのところとその先のR１つで1ステップ
    if &c[..1] == "W" && &c[1..2] == "R" {
        for i in 0..n {
            // Rが終わったらフラグを立てる
            if c.char_indices().nth(i).unwrap().1 == 'R' {
                if flag {
                    ans += 1
                }
                continue;
            } else {
                flag = true;
            }
        }
    } else {
        for i in 0..n {
            if c.char_indices().nth(i).unwrap().1 == 'R' {
                ans += 1
            }
        }
    }

    println!("{}", if flag { ans - 1 } else { ans });
}