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: Wrong Answer)

#include <stdio.h>

int main(void)
{
  int n,k;
  scanf("%d%d", &n, &k);

  char s[n+1];
  scanf("%s", s);

  int shakutori_array[n];
  
  int i;
  int j = 0;

  for(i = 0; i < n; i++) {
    if(j >= n) {
      shakutori_array[i] = -1;
      break;
    }
    else {
      shakutori_array[i] = 0;
    }

    if(i % 2 == 0) {
      while(s[j] == 'L') {
        shakutori_array[i]++;
        j++;
      }
    }
    else {
      while(s[j] == 'R') {
        shakutori_array[i]++;
        j++;
      }
    }
  }

  if(i <= 2 * k) {
    printf("%d\n", n-1);
  }
  else {
    int sum = 0;
    for(int m = 0; m <= 2 * k; m++) {
      sum += shakutori_array[m];
    }

    for(int m = 0; m < (i - 2 * k); m++) {
      if(shakutori_array[m] < shakutori_array[m + 2 * k]) {
        sum -= shakutori_array[m];
        sum += shakutori_array[m + 2 * k];
      }
    }

    printf("%d\n", sum - 1);
  }


  return 0;
}