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 Kotlin (1.3.71) (judge verdict: Time Limit Exceeded)

import java.util.*
fun main(args: Array<String>) {
    val reader = Scanner(System.`in`)
    var N = reader.nextLong()
    var str = reader.next()
    var start = 0
    var end = str.length - 1
    var findW = false
    var findR = false
    var count = 0
    do{
        if(str[start] == 'W'){
            findW = true
        }
        else {
            start += 1
        }

        if(str[end] == 'R'){
            findR = true
        }
        else {
            end -= 1
        }

        if(findW && findR){
            count += 1
            findW = false
            findR = false
        }
    } while(start  < end)
    
    println(count)
}