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 Ruby (2.3.3) (judge verdict: Wrong Answer)

def array(s, k)
  count = 0
  beforeLr = ""
  change = []
  s.each.with_index do |lr, i|
    if beforeLr == lr
      count += 1
      if k > 0
        change << i
        count += 2
        k -= 1
      end
    end
    if count > s.length - 1
      count = s.length - 1
    end
    beforeLr = lr
  end

  change.each do |ch|
    s[ch],s[ch - 1] = s[ch] == "L" ? "R" : "L"
  end

  return s, k, count
end

def onepoint(s,k,count)
  beforeLr = ""
  s.each.with_index do |lr, i|
    if beforeLr != lr
      if k > 0
        count += 1
        k -= 1
      end
    end
    if count > s.length - 1
      count = s.length - 1
    end
    beforeLr = lr
  end
  return s, k, count
end

n,k = gets.to_s.split.map(&:to_i)
s = gets.to_s.chomp.split("")
s, k, count = array(s, k)
s, k, count = onepoint(s, k, count)
puts count