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: Runtime Error)

//ABC140 D - Face Produces Unhappiness
//20190907

using System;
using System.Linq;
using System.Collections.Generic;

namespace D
{
    class Program
    {
        static void Main(string[] args)
        {
            int[] INPUT = Console.ReadLine().Split().Select(int.Parse).ToArray();
            int N = INPUT[0];
            int K = INPUT[1];


            string Sinput = Console.ReadLine();
            string [] S = new string[N];
            for(int i=0;i<N;i++)
            {
                S[i] = Sinput[i].ToString();
            }

            //計算
            // //尺取法で、最長区間を探して、その区間を反転する。
            for (int i = 0; i < K; i++)
            {
                int l = 0;
                int r = 1;
                int leftSide = 0, rightSide = 1;
                int max = 1;

                while (r < N - 1)
                {
                    while (S[r] == S[r + 1])
                    {
                        r++;
                    }

                    if (r - l > max)
                    {
                        leftSide = l;
                        rightSide = r;
                        max = r - l;
                    }

                    l = r;
                    r += 1;
                }


                var output = new List<string>();
                for (int j = leftSide; j < max; j++)
                {
                    output.Add(S[j].ToString());
                }
                output.Reverse();

                for (int j = leftSide; j < max; j++)
                {
                    S[i] = output[j];
                }
            }

            for (int i = 0; i < 1; i++)
            {

                int l = 0;
                int r = 1;
                int leftSide = 0, rightSide = 1;
                int max = 1;

                while (r < N - 1)
                {
                    while (S[r] == S[r + 1])
                    {
                        r++;
                    }

                    if (r - l > max)
                    {
                        leftSide = l;
                        rightSide = r;
                        max = r - l;
                    }

                    l = r;
                    r += 1;
                }

                Console.WriteLine(max);
            }


        }
    }
}