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)

# -*- coding: utf-8 -*-


def main():
    # See:
    # https://www.youtube.com/watch?v=VSeggcnxwrc

    # KeyInsight
    # L, Rでまとめる
    # 1回の操作で幸福度が最大で2増える
    # 操作したときの内部のスコアが変化しない
    # 例外的に，すべて同じ文字の場合は変化せず，ランレングス圧縮したときに2グループのときは1増える
    # min(理論値n - 1, 初期スコア + 2 * k回)

    # つまづきポイント
    # 反転操作で，スコアが変化する/変化しない部分を明らかにする
    # 操作が1回の場合を考えて，多数回に展開
    # スコアを増やす=異なるグループを減らす

    # どうしたら気がつけそうか?
    # 同じ要素をまとめる
    # 操作を言い換える，同じことを違う言葉で表現する/異なる視点で見る
    # 理論値VS初期値 + 操作による変化を比べる

    n, k = map(int, input().split())
    s = input()
    score = 0

    for i in range(n - 1):
        if s[i] == s[i + 1]:
            score += 1

    print(min(n - 1, score + 2 * k))


if __name__ == '__main__':
    main()