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 <iostream>
#include <string>
using namespace std;

int main()
{
  int n, k;
  cin >> n;
  cin >> k;
  string s;
  cin >> s;
  
  int gap = 0;
  for (int i = 0; i < n - 1; ++i)
  {
    if (s[i] == 'R' && s[i + 1] == 'L') ++gap;
  }
  
  int eGap = 0;
  if (s[0] == 'L') ++eGap;
  if (s[n-1] == 'R') ++eGap;
  
  int unhappy;
  if (k < gap)
  {
    unhappy = 2 * (gap - k) + eGap;
  }
  else if (k == gap)
  {
    unhappy = eGap == 0 ? 1 : eGap;
  }
  else
  {
    unhappy = 1;
  }
  cout << (n - unhappy) << endl;
  
  return 0;
}