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 PHP7 (7.0.15) (judge verdict: Accepted)

<?php
  list($n, $k) = explode(" ", trim(fgets(STDIN)));

  list($s) = explode(" ", trim(fgets(STDIN)));

  if($n <= 2){
    echo $n-1;
    exit;
  }

  $ans = 0;

  if($s[0]=='R' && $s[1]=='R'){
    $ans++;
  }
  if($s[$n-1] == 'L' && $s[$n-2] == 'L'){
    $ans++;
  }

  for ($i=1; $i<$n-1; $i++) {  
    if(($s[$i] == 'R' && $s[$i+1] == 'R') ||
       ($s[$i] == 'L' && $s[$i-1] == 'L')){
      $ans++;
    }
  }

  $fukou = $n - 1 - $ans;

  $kou = $fukou / 2;

  if($k >= $kou){
    echo $n-1;
    exit;
  }else{
    echo $ans + $k * 2;
    exit;
  }