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 java.io.StringReader;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Deque;
import java.util.List;
import java.util.Scanner;

public class Main {
	// LRRLRLRRLRLLR
	// LLLRLRLLRLRRR
	// LLLLRLRRLRRRR
	// LLLLLRLLRRRRR
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int k = sc.nextInt();
		sc.nextLine();
		List<Boolean> list = new ArrayList<Boolean>();

		int count = 0;
		for (String s : sc.nextLine().split("")) {
			if ("L".equals(s)) {
				list.add(Boolean.TRUE);
			} else {
				list.add(Boolean.FALSE);
			}
			count++;
		}
		int size = list.size();
		boolean l = list.get(0);
		boolean r = list.get(size - 1);
		for (int i = 0; i < k; i++) {
			// 一番左の異なるやつ
			int s = list.indexOf(!l);
			// 一番右の異なるやつ
			int e = list.lastIndexOf(!r);
			if (e < s) {
				e = size;
				for (int j = s; j <= e; j++) {
					list.set(j, !list.get(j));
				}
				break;
			}
			for (int j = s; j <= e; j++) {
				list.set(j, !list.get(j));
			}
		}
		boolean bb = list.get(0);
		int ans = 0;
		for (int i = 1; i < size; i++) {
			boolean b = list.get(i);
			if (bb == b) {
				ans++;
			}
			bb = b;
		}
		System.out.println(ans);
	}
}