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::{fastout, input};

#[fastout]
fn main() {
    input! {n: usize,s: String};
    let s: String = s;

    let mut v: Vec<char> = Vec::new();
    let mut count = 0;
    let mut red_count = 0;
    let mut index = 0;
    let mut red_d = 0;

    for c in s.as_str().chars() {
        if c == 'R' {
            red_count += 1;
        }

        v.push(c);
    }

    if red_count == n {
        println!("0")
    } else {
        while index < n {
            if red_d == red_count - 1 {
                // 最後の一個を消せばいい時
                println!("{}", count + 1);
                break;
            } else {
                // Wと後ろから見て最初に見つかったRを入れ替える
                if v[index] == 'W' {
                    //
                    for i in index..n {
                        if v[n - i - 1] == 'R' {
                            v[index] = 'R';
                            v[n - i - 1] = 'W';

                            red_d += 1;
                            index += 1;
                            count += 1;
                            break;
                        }
                    }
                //
                } else {
                    red_d += 1;
                    index += 1;
                }
            }
        }
    }
}