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;
using System.Collections.Generic;
using System.Diagnostics;
using System.IO;
using System.Linq;
using System.Text;

class Myon
{
    public Myon() { }
    public static int Main()
    {
        cin = new Scanner();
        new Myon().calc();
        return 0;
    }

    static Scanner cin;

    // メンバ変数---
    int N;
    int K;
    string[] S;
    // -------------

    void calc()
    {
        N = cin.nextInt();
        K = cin.nextInt();
        string s = cin.next();
        S = new string[N];

        for (var i = 0;i < N;i++)
        {
            S[i] = s[i].ToString();
        }

        int p = 0;
        for(int i = 0;i < N - 1; i++)
        {
            if(S[i] == S[i + 1])
            {
                p++;
            }
        }
        int ans = Math.Min(p+2*K, N - 1);
        write(ans);

    }

    void write(string v)
    {
        Console.WriteLine(v);
    }

    void write(int v)
    {
        Console.WriteLine(v);
    }

    void write(double v)
    {
        Console.WriteLine(v);
    }
    void write(long v)
    {
        Console.WriteLine(v);
    }

    int Pow(int x, int y)
    {
        return (int)Math.Pow(double.Parse(x.ToString()), double.Parse(y.ToString()));
    }

    double Pow(double x, double y)
    {
        return Math.Pow(x, y);
    }

    class Scanner
    {
        string[] s;
        int i;

        char[] cs = new char[] { ' ' };

        public Scanner()
        {
            s = new string[0];
            i = 0;
        }

        public string next()
        {
            if (i < s.Length) return s[i++];
            string st = Console.ReadLine();
            while (st == "") st = Console.ReadLine();
            s = st.Split(cs, StringSplitOptions.RemoveEmptyEntries);
            if (s.Length == 0) return next();
            i = 0;
            return s[i++];
        }

        public int nextInt()
        {
            return int.Parse(next());
        }
        public int[] ArrayInt(int N, int add = 0)
        {
            int[] Array = new int[N];
            for (int i = 0; i < N; i++)
            {
                Array[i] = nextInt() + add;
            }
            return Array;
        }

        public long nextLong()
        {
            return long.Parse(next());
        }

        public long[] ArrayLong(int N, long add = 0)
        {
            long[] Array = new long[N];
            for (int i = 0; i < N; i++)
            {
                Array[i] = nextLong() + add;
            }
            return Array;
        }

        public double nextDouble()
        {
            return double.Parse(next());
        }


        public double[] ArrayDouble(int N, double add = 0)
        {
            double[] Array = new double[N];
            for (int i = 0; i < N; i++)
            {
                Array[i] = nextDouble() + add;
            }
            return Array;
        }
    }
}