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: Time Limit Exceeded)

N, K = gets.split.map(&:to_i)
S = gets.chomp

def try(s, l, r)
  ret = s.dup
  ret[l..r] = ret[l..r].reverse.tr("LR", "RL")
  ret
end

def count(s)
  ret = 0
  prev_c = nil
  for i in 0...s.length
    c = s[i]
    if c == prev_c
      ret += 1
    end
    prev_c = c
  end
  ret
end

def make_cands(s)
  ret = []
  l = 0; r = 1; x = 0
  loop do
    #p l: l, r: r, x: x, s: s, slr: s[l..r]
    break if r >= s.length
    if s[r] == s[l]
      r += 1
    else
      s_ = try(s, l, r-1)
      #p l: l, r: r, c: count(s_), slr: s[l..r-1]
      item = [count(s_), s_, l, r-1]
      ret << item

      if x < 1 && !(r-l == 1)
        s_ = try(s, l, r)
        #p l: l, r: r, c: count(s_), slr: s[l..r]
        item = [count(s_), s_, l, r]
        ret << item
      end
      if x == 1
        c = s[l]
        l += 1 while s[l] == c
        r = l + 1
        x = 0
      else
        r += 1
        x += 1
      end
    end
  end
  s_ = try(s, l, r-1)
  ret << [count(s_), s_, l, r-1]
  ret
end
  
s = S; c = nil
K.times do
  cands = make_cands(s)
  m = cands.map{|n, _, _, _| n}.max
  c, s, l, r = cands.select{|n, _, _, _| n == m}
                    .max_by{|n, _, l, r| r-l}
  #p c: c, s: s, l: l, r: r
end
p c