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 static System.Console;
using static System.Math;
using System.IO;
using System.Collections.Generic;
namespace I_am_the_green_coder{
    class Inonoa{
        static void Main(){
            long[] nk = ILongs();
            char[] s = IChars();

            long change = 0;

            for(int i=0;i<nk[0]-1;i++){
                if(s[i]!=s[i+1])change ++;
            }
            long cantchange = Max(change - 2 * nk[1],0);

            WriteLine(nk[0]-1-cantchange);
        }

        static readonly long MOD = 1000000007;

        ///<summary>a_1, a_2, a_3, ...みたいなの</summary>
        static long[] ILongs(){
            string[] s = ReadLine().Split();
            long[] re = new long[s.Length];
            for(long i=0;i<s.Length;i++){
                re[i] = long.Parse(s[i]);
            }
            return re;
        }

        ///<summary>xooxoxoxxoxo...みたいなの</summary>
        static char[] IChars(){
            return ReadLine().ToCharArray();
        }

        ///<summary>n m とかをちょっとはやくかくだけ</summary>
        static string[] IStrings(){
            return ReadLine().Split();
        }

        ///<summary>n行の入力を取ったり</summary>
        static long[][] ILongLongs(long n){
            long[][] re = new long[n][];
            for(long i=0;i<n;i++){
                string[] s = ReadLine().Split();
                re[i] = new long[s.Length];
                for(long j=0;j<s.Length;j++){
                    re[i][j] = long.Parse(s[j]);
                }
            }
            return re;
        }

        ///<summary>n行の文字列を取ったり</summary>
        static char[][] ICharsChars(long n){
            char[][] re = new char[n][];
            for(long i=0;i<n;i++){
                re[i] = ReadLine().ToCharArray();
            }
            return re;
        }
    }
}