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: Accepted)

import java.util.*
import java.io.PrintWriter



fun main(args: Array<String>){
	val cin = Scanner(System.`in`)
    val cout = PrintWriter(System.`out`)
    val n = cin.nextInt()
    val k = cin.nextInt()
    cin.nextLine()
    val s = "R" + cin.nextLine() + "L"

    // var cut = 0
    var unhappy = 0
    var ans : Int
    for (i in 1..n) {
        if ((s[i-1] == 'R' && s[i] == 'L') || (s[i] == 'R' && s[i+1] == 'L')) {
            unhappy += 1
        }
    }
    if (unhappy > k*2) {
        ans = n - (unhappy - k*2)
    } else {
        ans = n - 1
    }

    cout.println(ans)
    
    cout.flush()
}


fun next() = readLine()!!
fun nextInt() = next().toInt()
fun nextLong() = next().toLong()
fun nextDouble() = next().toDouble()
fun listOfString() = next().split(" ")
fun listOfInt() =listOfString().map { it.toInt() }
fun listOfLong() =listOfString().map { it.toLong() }
fun listOfDouble() =listOfString().map { it.toDouble() }