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: Accepted)

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

first = 1
while first < n && k > 0
    if s[first - 1] != s[first] then
        last = first
        while last < n - 1
            if s[last] == s[first] && s[last + 1] == s[first - 1] then
                break
            end
            last += 1
        end
        for i in first.upto(last)
            if s[i] == 'L' then
                s[i] = 'R'
            else
                s[i] = 'L'
            end
        end
        k -= 1
    end
    first += 1
end

count = 0
for i in 1.upto(n - 1)
    if s[i - 1] != s[i] then
        count += 1
    end
end

puts n - 1 - count