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: Wrong Answer)

N,K = map(int,input().split(' '))
S = input()
previous_idx = 0
if S[0] == 'L':
    search_s = 'R'
    non_search_s = 'L'
else:
    search_s = 'L'
    non_search_s = 'R'

continue_flag = False
idx = 0
lst_idx =  -5
new_S = ''
while K != 0 and idx != N:
    if S[idx] == search_s:
        if not continue_flag:
            continue_flag = True
        new_S += non_search_s
    else:
        if continue_flag:
            continue_flag = False
            K -= 1
            last_idx = idx
        new_S += S[idx]
    idx += 1
new_S += S[idx:N]
print(new_S)

count_s = 0
previous_s = new_S[0]
for i in range(1,N):
    if previous_s == new_S[i]:
        count_s +=1
    else:
        previous_s = new_S[i]
print(count_s)