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: Accepted)

#include <bits/stdc++.h>
#define SORT(x) sort(x.begin(),x.end())
#define ALL(x) x.begin(),x.end()
#define rep(i, n) for(int i = 0; i < n; i++)
#define reps(i, m, n) for(int i = m; i < n; i++)
#define repr(i, m, n) for(int i = m; i >= n; i--)
#define INF 1000000007 //1000000000000000003
#define PI (acos(-1))
#define MOD 1000000007
using namespace std;
typedef long long ll;



void Main(){

  int N,K;
  cin >> N>>K;

  string S;
  cin>>S;

  int ct=0;
  rep(i,N-1){
    ct+= S[i]==S[i+1];
  }

  cout<< min(ct+2*K,N-1)<<"\n";

}


//-----------------------------------
int main(){
  cin.tie(nullptr);
	ios_base::sync_with_stdio(false);
	cout << fixed << setprecision(15);
	Main();
}
//-----------------------------------