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 PyPy3 (7.3.0) (judge verdict: Wrong Answer)

n,k=map(int,input().split())
s=list(input())

RL=0
LR=0

fo=0

if n<3:
  if n==1:
    print(0)
  else:
    print(1)
    
else:
  for i in range(n):
    if s[i]=='R' and i<n-1:
      if s[i+1]=='R':
        fo+=1
    if s[i]=='L' and 0<i:
      if s[i-1]=='L':
        fo+=1
        
    if i<n-1:
      if s[i]=='R' and s[i+1]=='L':
        RL+=1
      elif s[i]=='L' and s[i+1]=='R':
        LR+=1
        
  si=min(k,RL,LR)
  no=k-si
  LR_rest=LR-si
  RL_rest=RL-si
  if no==0:
    ans=fo+si*2
  else:
    ans=fo+si*2
    if s[0]=='L':
      if LR_rest>0:
        ans+=1
        LR_rest-=1
        no-=1
    if no>0:
      if s[-1]=='R':
        if LR_rest>0:
          ans+=1
  print(ans)