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)

n, k = gets.chomp.split.map(&:to_i)
s = gets.chomp.split('')

def count(a)
  sum = 0
  asize = a.size
  asize.times do |i|
    sum += 1 if (a[i] == 'L' && i != 0 && a[i - 1] == 'L') || (a[i] == 'R' && i != asize - 1 && a[i + 1] == 'R')
  end
  sum
end

def count_reverse(a)
  sum = 0
  (a.size - 1).times do |i|
    sum += 1 if a[i] != a[i + 1]
  end
  sum
end

reverse = count_reverse(s)

max = count(s)
k.times do
  reverse -= 2 if reverse >= 2
  reverse -= 1 if reverse == 1
  break if reverse == 0
end
p n - reverse - 1