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.split.map(&:to_i)
face = gets.chomp.split("")
happy = 0
count = 0
for i in 1..n-1
  if face[i] == face[i-1]
    happy += 1
  else
    count += 1
  end
end
min = 1
max = 2
for i in 1..k
  check = face[min-1]
  for j in min..n-1
    if face[j-1] != face[j] && face[j] != check
      min = j
      break
    end
  end
  check = face[-max+1]
  for k in max..n
    if face[-k+1] != face[-k] && face[-k] != check
      max = k
      break
    end
  end
  if min > n-max
    break
  elsif min == n-max
    happy += 1
    break
  end
  happy += 2
  min += 1
  max += 1
end
if count == 1
  happy += 1
end
puts happy