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 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);
    }

    for i in 0..red_count {
        if v[i] == 'W' {
            count += 1;
        }
    }

    println!("{}", count);

    //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;
    //            }
    //        }
    //    }
    //}
}