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: Wrong Answer)

import re

def nchars(s, n):
    """文字列 s に、同じ文字が n 個以上連続している部分文字列を見つける
    """
    assert n > 0
    reg = re.compile("(.)\\1{%d,}" % (n - 1))  # カンマを取ると n 個ちょうどになる
    while True:
        m = reg.search(s)
        if not m:
            break
        yield m.group(0)
        s = s[m.end():]

def get_largest(arr, int):
    if int == 0:
        return arr
    sum = 0
    index = 0
    len = len(range)
    for i in range(len):
        if i < len - 2:
            if sum < arr[i] + arr[i+1] + arr[i+2]:
                sum = arr[i] + arr[i+1] + arr[i+2]
                index = i
    int -= 1


def main():
    [N, K] = [int(number) for number in input().split()]
    S = input()
    s = [len(rl) for rl in list(nchars(S,1))]
    i = K
    while (i > 0):
        len_s = len(s)
        if len_s < 3:
            break
        three_range_arr = [s[j] + s[j+1] + s[j+2] for j in range(len_s) if j < len_s-2]
        max_tra = max(three_range_arr)
        index = three_range_arr.index(max_tra)
        if index == 0:
            ss = [max_tra]
            ss.extend(s[index+3:len_s])
        elif index == len_s - 3:
            ss = s[0:index-1]
            ss.extend([max_tra])
        else:
            ss = s[0:index-1]
            ss.extend([max_tra])
            ss.extend(s[index+3:len_s])
        #s = s[0:index-2].append(max_tra)#.extend(s[index+1, len_s-1])
        i-=1
        s=ss

    if i > 0:
        print(N-1)
    else:
        print(sum(s)-len(s))
    #ans_arr = get_largest(s, K)

if __name__ == '__main__':
    main()