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 (2.4.0) (judge verdict: Accepted)

# input = sys.stdin.readline
from bisect import *
from collections import *
from heapq import *
# import functools
# import itertools
# import math
N,K=map(int,input().split())
S=input()
if N==1:
    print(0)
    exit()

_K=K
_S=list(S)

pre=_S[0]
flag=0
for i in range(1,N):

        if (_K>0 or flag==1 )and pre!=_S[i]:
            flag=1
            _S[i]=pre
        elif pre==_S[i] and flag==1:
            _K-=1
            flag=0

count=0
if _S[0]=="R" and _S[1]=="R":
    count+=1
if _S[-1]=="L" and _S[-2]=="L":
    count+=1

for i in range(1,N-1):
    if _S[i-1]=="L" and _S[i]=="L"  or _S[i]=="R" and _S[i+1]=="R":
        count+=1

print(count)