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 abc140_d
{
  class Program
  {
    static void Main(string[] args)
    {
      var n = CIn.ReadInt();
      var k = CIn.ReadInt();
      var s = CIn.Read().ToCharArray();

      var sum = 0;
      var lr = 0;
      var rl = 0;

      for (var i = 1; i < n; i++)
      {
        if (s[i - 1] == s[i])
        {
          sum++;
          continue;
        }

        if (s[i] == 'R')
        {
          lr++;
        }
        else
        {
          rl++;
        }
      }

      var min = Math.Min(k, Math.Min(lr, rl));
      var ans = sum + min * 2 + (min != k ? Math.Abs(lr - rl) : 0);

      Console.WriteLine(ans);
    }
  }

  class CIn
  {
    private static readonly Queue<string> fieldBuffer = new Queue<string>();

    public static string Read()
    {
      if (fieldBuffer.Count == 0)
      {
        foreach (var field in Console.ReadLine().Split())
        {
          fieldBuffer.Enqueue(field);
        }
      }

      return fieldBuffer.Dequeue();
    }

    public static int ReadInt() => int.Parse(Read());
    public static long ReadLong() => long.Parse(Read());
    public static double ReadDouble() => double.Parse(Read());

    public static string[] ReadStringArray(long n)
    {
      var array = new string[n];

      for (var i = 0; i < n; i++)
      {
        array[i] = Read();
      }

      return array;
    }

    public static int[] ReadIntArray(long n)
    {
      var array = new int[n];

      for (var i = 0; i < n; i++)
      {
        array[i] = ReadInt();
      }

      return array;
    }

    public static long[] ReadLongArray(long n)
    {
      var array = new long[n];

      for (var i = 0; i < n; i++)
      {
        array[i] = ReadLong();
      }

      return array;
    }
    public static double[] ReadDoubleArray(long n)
    {
      var array = new double[n];

      for (var i = 0; i < n; i++)
      {
        array[i] = ReadDouble();
      }

      return array;
    }
  }
}