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 5.4.1) (judge verdict: Accepted)

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MIN(a, b) (a<b?a:b)

int cmp(const void *a, const void *b) {
	return *(int*)b - *(int*)a;
}

int main(int argc, char const* argv[])
{
	int n, i;
	int k;
	char s[100010];
	int l=0, r=0;
	int ans=0;
	scanf("%d%d%s",&n, &k, s+1);
	s[0] = '_';
	for(i=1; i<=n; i++) {
		if(s[i]=='L') {
			if(s[i-1]=='L') {
				ans++;
			} else {
				l++;
			}
		} else if(s[i]=='R') {
			if(s[i-1]=='R') {
				ans++;
			} else {
				r++;
			}
		}
	}
	int temp = MIN(l, r);
	if(temp > k) {
		ans += k*2;
	} else {
		ans = n-1;
	}
	printf("%d\n", ans);
	
	return 0;
}