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 Java8 (OpenJDK 1.8.0) (judge verdict: Runtime Error)

import javax.swing.text.MutableAttributeSet;
import java.util.*;

public class Main {

    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int k = sc.nextInt();

        String s = sc.next();
        List<Integer> list = new ArrayList<>();
        int index = 0;
        int ans = 0;
        for (int i = 1; i < n; i++) {
            while (s.charAt(i) == s.charAt(i - 1)) {
                i++;
                if(i >= n){
                    break;
                }
            }
            list.add(i - index);
            ans += i - index - 1;
            index = i;
        }
        if(s.charAt(n-1) != s.charAt(n-2)){
            list.add(1);
        }

        if(list.size() % 2 == 0){
            if(((list.size()/2)-1) >= k){
                ans += k * 2;
            } else {
                ans = n-1;
            }

        } else {
            ans += Math.min(list.size()-1, k*2);
        }
        System.out.println(ans);
    }
}