Problem:
Score : 400 points

Problem Statement

There are N people standing in a queue from west to east.

Given is a string S of length N representing the directions of the people.
The i-th person from the west is facing west if the i-th character of S is L, and east if that character of S is R.

A person is happy if the person in front of him/her is facing the same direction.
If no person is standing in front of a person, however, he/she is not happy.

You can perform the following operation any number of times between 0 and K (inclusive):

Operation: Choose integers l and r such that 1 \leq l \leq r \leq N, and rotate by 180 degrees the part of the queue: the l-th, (l+1)-th, ..., r-th persons. That is, for each i = 0, 1, ..., r-l, the (l + i)-th person from the west will stand the (r - i)-th from the west after the operation, facing east if he/she is facing west now, and vice versa.

What is the maximum possible number of happy people you can have?

Constraints

N is an integer satisfying 1 \leq N \leq 10^5.

K is an integer satisfying 1 \leq K \leq 10^5.

|S| = N

Each character of S is L or R.

Input

Input is given from Standard Input in the following format:

N K
S

Output

Print the maximum possible number of happy people after at most K operations.

Sample Input 1

6 1
LRLRRL

Sample Output 1

3

If we choose (l, r) = (2, 5), we have LLLRLL, where the 2-nd, 3-rd, and 6-th persons from the west are happy.

Sample Input 2

13 3
LRRLRLRRLRLLR

Sample Output 2

9

Sample Input 3

10 1
LLLLLRRRRR

Sample Output 3

9

Sample Input 4

9 2
RRRLRLRLL

Sample Output 4

7

Solution in Kotlin (1.0.0) (judge verdict: Wrong Answer)

import java.util.*

fun main(args: Array<String>) {
    val scanner = Scanner(System.`in`)
    val n = scanner.nextInt()
    val k = scanner.nextInt()
    val s = scanner.next()
    val s2 = s.split("")
    var left = 0
    var right = 0
    var check = 0
    var count = 0
    var count2 = 0
    var count3 = 0
    for(i in 0..n-1){
        if(check === 1) {
            if (s[i] == 'L' && i > 0) {
                if(s[i-1] == 'L') {
                    count++
                    count3++
                }
                else count3 = 0


            } else if (s[i] == 'R' && i < n-1) {
                if(s[i+1] == 'R') {
                    count++
                    count3++
                }
                else count3 = 0

            }
        }
        else{
            if (s[i] == 'L' && i > 0) {
                if(s[i-1] == 'L') count2++
                else check++

            } else if (s[i] == 'R' && i < n-1) {
                if(s[i+1] == 'R') count2++
                else check++

            }
        }
    }
    val num = n-1-count-count2
    if(num >= k) println(count+count2+2*k)
    else if(count2 === 0 && count3 === 0) println(count+count2+2*num)
    else {
        if(k-num >= 2) {
            if (count2 > 0) count++
            if (count3 > 0) count++
            println(count + count2 + 2 * num)
        }
        else if(k-num === 1) {
            println(count + count2 + 2 * num+1)
        }
    }
}