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.*

class Scanner(`in`: java.io.InputStream) {
    private val reader = java.io.BufferedReader(java.io.InputStreamReader(`in`))
    private var tokenizer: StringTokenizer? = null

    fun next(): String {
        if (tokenizer?.hasMoreTokens() != true) {
            tokenizer = StringTokenizer(reader.readLine())
        }
        return tokenizer!!.nextToken()
    }

    fun nextLine(): String {
        return if (tokenizer?.hasMoreTokens() != true) {
            reader.readLine()
        } else tokenizer!!.nextToken("\n")
    }
}

fun main(args: Array<String>) {
    val pw = java.io.PrintWriter(System.out)
    Problem.solve(Scanner(System.`in`), pw)
    pw.flush()
}

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object Problem {
    fun solve(sc: Scanner, pw: java.io.PrintWriter) {
        val N = sc.next().toInt()
        val K = sc.next().toInt()
        val S = sc.next().toCharArray()

        var ans = 0
        var n = 0
        for (i in 1 until N) {
            if (S[i] != S[i - 1]) {
                n++
            } else {
                ans++
            }
        }

        var k = K
        while (n > 1 && k > 0) {
            ans += 2
            n -= 2
            k--
        }

        if (k > 0 && n == 1 && (S[0] == 'L' || S[N - 1] == 'R')) {
            ans++
        }

        pw.println(ans)
    }
}