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 Python (3.4.3) (judge verdict: Accepted)

def new_list(s_list):
    s_list_new = [s_list[0]]
    for i, s in enumerate(s_list):
        if i >= 1:
            if s != s_list[i - 1]:
                s_list_new.append(s)
    return s_list_new

if __name__ == "__main__":

    count = 0
    n, k = map(int, input().split())
    stext = str(input())
    s_list = [s for s in stext]
    s_list = new_list(s_list)
    cnt_l = s_list.count('L')
    cnt_r = s_list.count('R')
    if cnt_l == cnt_r:
        if k >= cnt_l:
            count = 1
        else:
            count = 2 * (cnt_l - k)
    else:
        cnt_lr = min(cnt_l, cnt_r)
        if k >= cnt_lr:
            count = 1
        else:
            count = 2 * (cnt_lr - k) + 1
                
    print('{0}'.format(n - count))