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;

namespace abc140_d
{
    static class Program
    {
        static void Main(string[] args)
        {
            var items = Console.ReadLine().ToIntArray();
            // 人数
            var n = items[0];
            // 回転させる回数
            var k = items[1];
            // 向いている方向
            var s = Console.ReadLine();

            // 向いている方向が同じ人たちの列が最も長くなるところを回転させたい
            // →よくよく考えたら、(両端以外なら)どの場所を入れ替えても、幸福の増加量は+2では？
            // →一個飛ばしで、前から順々に入れ替えていくのが一番いい

            // もともとの合計値
            var sum = 0L;
            // 直前のグループの末尾インデックス
            var prev = 0;
            // グループの総数
            var count = 0;
            for (int i = 1; i < s.Length; i++)
            {
                if (s[i - 1] != s[i])
                {
                    sum += i - prev - 1;
                    prev = i;
                    count++;
                }
            }
            sum += s.Length - prev - 1;
            count++;

            // 各グループ数ごとの、最大回転数は下記の通り

            // グループ数 1 : 0回 o +0
            // グループ数 2 : 1回 ox +1
            // グループ数 3 : 1回 oxo +2
            // グループ数 4 : 2回 oxox +2+1
            // グループ数 5 : 2回 oxoxo
            // グループ数 6 : 3回 oxoxox
            // グループ数 7 : 3回 oxoxoxo

            var max = count / 2;
            if (k < max)
            {
                sum += k * 2;
            }
            else
            {
                sum += max * 2;
                // グループが偶数だったら、両端を回転するので、増加量が-1になることに注意
                if (count % 2 == 0)
                {
                    sum -= 1;
                }
            }

            Console.WriteLine($"{sum}");
        }

        static int ToInteger(this string s)
        {
            return int.Parse(s);
        }

        static int[] ToIntArray(this string s, char separator = ' ')
        {
            return s.Split(separator).Select(m => m.ToInteger()).ToArray();
        }
    }
}