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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
 
int main() {
    int n,k,i,j,ans,co;
    co=0;
    cin>>n>>k;
    char a[n];
    rep(i,n){
      cin>>a[i];
    }
    ans=0;
   rep(i,n-1){
     if(a[i]=='R'&&a[i+1]=='R'){
         ans++;
     }
   }
   for(i=n-1;i>0;i--){
     if(a[i]=='L'&&a[i-1]=='L'){
         ans++;
     }
   }
   for(i=0;i<n-1;i++){
       if(a[i]=='L'&&a[i+1]=='R'){
        co++;
       }
       if(a[i]=='R'&&a[i+1]=='L'){
        co++;
       }
   }
   if(co+1-2*k>2){
   cout<<ans+k*2<<endl;
   }else if(co==0){
       cout<<ans<<endl;
   }
   else {
   cout<<ans+co*2-1<<endl;
   }
}