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: Wrong Answer)

using System;
using System.Text;
using System.Linq;
using System.Collections;
using System.Collections.Generic;
using static System.Console;
using static System.Math;

namespace AtCoder
{
    public class Program
    {
        public static void Main(string[] args)
        {
            new Program().Solve(new ConsoleInput(Console.In, ' '));
        }

        public void Solve(ConsoleInput cin)
        {
            var N = cin.ReadInt;
            var K = cin.ReadInt;
            var S = cin.Read;

            var chars = S.ToCharArray();
            var ans = 0;
            for (int i = 1; i < N; i++)
            {
                if (chars[i - 1] == chars[i])
                {
                    ans++;
                }
                else if (chars[i - 1] == 'L' && chars[i] == 'R')
                {
                    if (K > 0)
                    {
                        ans += 2;
                        K--;
                    }
                }
            }
            ans = Min(ans, N - 1);
            Console.WriteLine(ans);
        }
    }

    public class ConsoleInput
    {
        private readonly System.IO.TextReader _stream;
        private char _separator = ' ';
        private Queue<string> inputStream;
        public ConsoleInput(System.IO.TextReader stream, char separator = ' ')
        {
            this._separator = separator;
            this._stream = stream;
            inputStream = new Queue<string>();
        }
        public string Read
        {
            get
            {
                if (inputStream.Count != 0) return inputStream.Dequeue();
                string[] tmp = _stream.ReadLine().Split(_separator);
                for (int i = 0; i < tmp.Length; ++i)
                    inputStream.Enqueue(tmp[i]);
                return inputStream.Dequeue();
            }
        }
        public string ReadLine { get { return _stream.ReadLine(); }}
        public int ReadInt { get { return int.Parse(Read); }}
        public long ReadLong { get { return long.Parse(Read); }}
        public double ReadDouble { get { return double.Parse(Read); }}
        public string[] ReadStrArray(long N) { var ret = new string[N]; for (long i = 0; i < N; ++i) ret[i] = Read; return ret;}
        public int[] ReadIntArray(long N) { var ret = new int[N]; for (long i = 0; i < N; ++i) ret[i] = ReadInt; return ret;}
        public long[] ReadLongArray(long N) { var ret = new long[N]; for (long i = 0; i < N; ++i) ret[i] = ReadLong; return ret;}
    }
}