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];
  scanf("%s",&s);
  int two=0,one=0,ord=0;
  for(int i=0;i<n-1;i++){//元の幸福度をカウント
	if(s[i]==s[i+1])ord++;
  }
  for(int i=0;i<n;i++){
    if(k==0)break;
	if(s[i+1]!=s[i]){i++;
      while(s[i]==s[i+1]&&i<=n-1){
        s[i]=s[i-1];
        i++;
                      }
      if(i==n-1){
        one++;
      k--;
                }
      else if(s[i]!=s[i+1]){
        s[i]=s[i+1];
        two++;//両端が異なる要素数をカウント
        k--;
      }
       
    }
  }	
/*  if(k<two)printf("%d",ord+2*k);
  else if(k>two&&one+two<k)printf("%d",2*two+k-two);
  else printf("%d",n-1);*/
  printf("%d",2*two+one+ord);
  return 0;
}