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.util.*;
public class Main {

	public static void main(String[] args) {
		Scanner stdIn = new Scanner(System.in);
		int N = stdIn.nextInt();
		int K = stdIn.nextInt();
		String S = stdIn.next();
		char[] s = S.toCharArray();
		long sum = 0;
		int out = 0;
		int in = 0;
		
		if(s[0] == 'R' && s[1] == 'R') sum++;
		else if(s[0] == 'L' && s[1] == 'R') out++;
		else if(s[0] == 'R' && s[1] == 'L') in++;
		
		if(s[N-1] == 'L' && s[N-2] == 'L') sum++;
		for(int i = 1; i < N-1; i++) {
			if(s[i] == 'L' && s[i-1] == 'L') sum++;
			else if(s[i] == 'R' && s[i+1] == 'R') sum++;
			
			if(s[i] == 'L' && s[i+1] == 'R') out++;
			else if(s[i] == 'R' && s[i+1] == 'L') in++;
		}
		
		if(Math.min(out,in) >= K) {
			sum += K*2;
		}
		else {
			sum += Math.min(out,in)*2;
			if(out == in);
			else if(K-Math.min(out,in) == 1 || Math.max(out,in)-Math.min(out,in) == 1) {
				if(Math.max(out,in) == out && (s[0] == 'L' || s[N-1] == 'R')) sum++;
				else if(Math.max(out,in) == in && (s[0] == 'R' || s[N-1] == 'L')) sum++;
			}
			else {
				if(Math.max(out,in) == out) {
					if(s[0] == 'L')sum++;
					if(s[N-1] == 'R')sum++;
				}
				else if(Math.max(out,in) == in && s[0] == 'R') {
					if(s[0] == 'R')sum++;
					if(s[N-1] == 'L')sum++;
				}
			}
		}
		
		System.out.println(sum);

	}

}