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.Linq;
using System.Collections.Generic;

using E = System.Linq.Enumerable;

public class Program
{
    public static void Main()
    {
        var nk = Console.ReadLine().Split().Select(x => int.Parse(x)).ToArray();
        var n = nk[0];
        var k = nk[1];
        var s = Console.ReadLine().Trim();

        var happyness = 0;
        var m = 1;
        var l = 0;
        for (int i = 0; i < s.Length; ++i)
        {
            if (s[i] == s[l])
            {
                continue;
            }

            happyness += i - l - 1;
            l = i;
            ++m;
        }
        happyness += s.Length - l - 1;

        k = Math.Min(k, m / 2);
        happyness += k * 2 - (k * 2 == m ? 1 : 0);

        Console.WriteLine(happyness);
    }
}