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 (GCC 9.2.1) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
char s[200000];
int a[200000];
int main()
{
	int n, k;
	int d = 0;
	scanf("%d", &n);
	scanf("%d\n", &k);
	d = 0;
	gets(s);
	int i, j, c;
	int b;
	int l = 1;
	int ml[3] = { 0 };
	c = 0;

	b = 0;
	while (c < k)
	{
		i = 0;
		d = 1;
		b = 0;
		ml[0] = 0;
		ml[1] = 0;
		ml[2] = 0;
		j = 1;
		a[0] = 0;
		l = 0;
		while (i + 1 < n)
		{
			if (s[i] == s[i + 1])
			{
				if (j == 0)
				{
					j = 1;
					l++;
				}
				else
					l++;

			}
			else
			{
				a[d] = i;
				a[d + 1] = i + 1;
				if (b < 2)
				{
					b++;
					l++;
				}
				else
				{
	
					if (l > ml[0])
					{
						ml[0] = l;
						ml[1] = a[d - 4];
						ml[2] = a[d];
					}
					l = l - a[d - 3] + a[d - 5] + 1;
				}
				j = 0;
				d = d + 2;
			}
			i++;
		}
		if (b < 2)
		{
			d = 0;
			while (d < n - 1)
			{
				s[d + 1] = s[d];
				d++;
			}
			ml[0] = l;
		}
		else
		{
			d = ml[1];
			while (d < ml[2])
			{
				s[d + 1] = s[d];
				d++;
			}
		}
		c++;
	}
	d = 0;
	l = 0;
	while (d < n - 1)
	{
		if (s[d] == s[d + 1])
			l++;
		d++;
	}
	printf("%d\n", l);
	return 0;
}