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: Wrong Answer)

import java.util.Scanner;

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();

    char cur = s.charAt(0);
    int ans = 0;
    int change = 0;
    for (int i = 1; i < n; ++i) {
      if (cur == s.charAt(i)) {
        ++ans;
      } else {
        cur = s.charAt(i);
        ++change;
      }
    }

    if (change == 1) {
      ++ans;
    } else if (change != 0) {
      int count = 0;
      int i = 0;
      int j = n - 1;
      while (count < k) {
        int left = i++;
        int right = j--;
  
        while (i < n && s.charAt(left) == s.charAt(i)) ++i;
        while (j >= 0 && s.charAt(right) == s.charAt(j)) --j;
  
        if (i <= j) {
          ans += 2;
        } else {
          break;
        }
        ++count;
      }  
    }
    
    System.out.println(ans);

    sc.close(); 
  }
}