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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Collections.Generic;

namespace AtCoder.Abc140
{
    class QuestionD
    {
        public static void Main(string[] args)
        {
            // 整数の入力
            string[] input = Console.ReadLine().Split(' ');
            int n = int.Parse(input[0]);
            int k = int.Parse(input[1]);
            char[] s = Console.ReadLine().ToCharArray();

            if(n == 1)
            {
                Console.WriteLine(0);
                return;
            }

            //今の幸福度と、RLの並び、およびLRの並びをそれぞれ数える
            int score = 0;
            if(s[0] == 'R' && s[1] == 'R')
            {
                score++;
            }
            for(int i = 1; i < s.Length -1; i++)
            {
                if(s[i] == 'R')
                {
                    if (s[i + 1] == 'R')
                    {
                        score++;
                    }
                }
                else
                {
                    if (s[i - 1] == 'L')
                    {
                        score++;
                    }
                }
            }
            if (s[n-2] == 'L' && s[n-1] == 'L')
            {
                score++;
            }

            //操作を1度やるたびに、幸福な人が2人増える
            int result = score + (k * 2);
            //ただし、最大幸福度はn-1以上にはならない
            if(result >= n)
            {
                result = n - 1;
            }
            Console.WriteLine(result);
        }
    }
}