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: Accepted)

import java.util.*;
import static java.lang.Math.*;
import java.math.BigInteger;

public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);

		// 入力
		int n = sc.nextInt();
		int k = sc.nextInt();
		String s = sc.next();
		String c[] = s.split("");


		// 計算
		int happy = 0;
		int lr = 0;
		int rl = 0;
		for(int i = 0; i < n; i++){
		    /*
		    if(i == 0 && c[i].equals("L")){
		        rl++;
		    }
		    if(i == n - 1 && c[i].equals("R")){
		        rl++;
		    }*/
		    
		    if(i > 0){
		        if(c[i - 1].equals(c[i])){
		            happy++;
		        }else{
		            if(c[i].equals("R")){
		                lr++;
		            }else{
		                rl++;
		            }
		        }
		    }
		}
		int result = happy + min(min(lr, rl), k)*2;
		if(abs(lr - rl) > 0 && k - min(lr, rl) > 0) result++;

		// 出力
		System.out.println(result);
	}
}