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: Time Limit Exceeded)

import sys
import copy
import re
n, k = [int(i) for i in sys.stdin.readline().split()]
s = input()
def _reverse(text):
    res = text[::-1]
    res = ["R" if i == "L" else "L" for i in res]
    res = "".join(res)
    return res
"""
for i in range(k):
    left = s.find("R")
    left = 0 if left == -1 else left
    right = s.rfind("R") + 1

    s = s[:left] + _reverse(s[left:right]) + s[right:]
    #s = copy.deepcopy(s_new)
cnt = sum([len(i) - 1 for i in re.findall(r"R{2,}|L{2,}", s)])
s = copy.deepcopy(s_origi)
"""
for i in range(k):
    left = s.find("LR") + 1
    left = 0 if left == -1 else left
    right = s.rfind("RL")
    right = n if right == -1 else right + 1

    s = s[:left] + _reverse(s[left:right]) + s[right:]
    #s = copy.deepcopy(s_new)
cnt = sum([len(i) - 1 for i in re.findall(r"R{2,}|L{2,}", s)])
print(cnt)