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.*;

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

      int n = Integer.parseInt(sc.next());
      int k = Integer.parseInt(sc.next());
      String s = sc.next();

      char[] c = new char[n];

      int ans = 0;

      int numRL = 0;

      for(int i=0;i<n;i++){
        c[i] = s.charAt(i);
        if(i!=0 && c[i]==c[i-1]){
          ans++;
        }
        if(i!=0 && c[i-1]=='R' && c[i]=='L' ){
          numRL++;
        }
      }

      if(numRL>k){
        ans += 2*k;
      }
      else{
        ans = n -1;
        if(c[0]=='R' && c[0]=='R'){
          ans-=2;
        }
      }


      System.out.println(ans);


    }
}