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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <stdio.h>
int main(){
    int n,k,ans;
    int p=0;
    int q=0;
    scanf("%d %d",&n,&k);
  
    int S[n];
    char s[n];
    scanf("%s",s);
  
    for(int i=0;i<n;i++){
      if('L'==s[i]){
        S[i]=0;
      }
    }
    
    for(int i=0;i<n;i++){
        if(S[i]!=S[i+1]){
            p=p+1;
        }else{
          q=q+1;
        }
    }
    
    int t=k;
    if(t%2==1){
      t=t+1;
    }
  
    if(p<=t/2){
      ans=n-1;
    }else{
      ans=q+2*k;
    }
    
    printf("%d",ans);
    
    return 0;
}