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;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace AtCorder
{
	public class Program
	{
		static void Main(string[] args)
		{
			Scanner cin = new Scanner();
			int n = cin.Int();
			int k = cin.Int();
			string s = cin.Next();
			int cnt = 0;
			char last = 'R';
			for (int i = 0; i < n; i++) {
				if (s[i] == 'R') {
					if (i != n - 1 && s[i + 1] == 'R') {
						cnt++;
					}
				} else {
					if (i != 0 && s[i - 1] == 'L') {
						cnt++;
					}
				}
			}

			Console.WriteLine(Math.Min(n - 1, cnt + 2 * k));
		}
	}

	public class Scanner
	{
		private readonly char[] delimiter_ = new char[] { ' ' };
		private string[] buf_;
		private int index_;

		public Scanner()
		{
			buf_ = new string[0];
			index_ = 0;
		}

		public string Next()
		{
			if (index_ < buf_.Length) {
				return buf_[index_++];
			}

			string st = Console.ReadLine();
			while (st == "") {
				st = Console.ReadLine();
			}

			buf_ = st.Split(delimiter_, StringSplitOptions.RemoveEmptyEntries);
			if (buf_.Length == 0) {
				return Next();
			}

			index_ = 0;
			return buf_[index_++];
		}

		public int Int()
		{
			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] = Int() + add;
			}
			return Array;
		}

		public long Long()
		{
			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] = Long() + add;
			}
			return Array;
		}

		public double Double()
		{
			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] = Double() + add;
			}
			return Array;
		}
	}
}