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>
using namespace std;

typedef long long ll;

template<typename T>
void out(T x) { cout << x << endl; exit(0); }
#define watch(x) cout << (#x) << " is " << (x) << endl

const int maxn = 1e6 + 5;

int n,k;
string s;

int main() {
  ios_base::sync_with_stdio(false); cin.tie(0);  cout.tie(0);
  cin>>n>>k>>s;

  int diffs = 0;
  for (int i=0; i+1<n; i++) {
    if (s[i] != s[i+1]) diffs++;
  }

  int good = 0;
  for (int i=0; i<n; i++) {
    if (s[i] == 'L' && i > 0 && s[i-1]=='L') good++;
    if (s[i] == 'R' && i+1 < n && s[i+1]=='R') good++;
  }

  //watch(diffs);
  //watch(good);
  while (k>0 && diffs>0) {
    int dx = min(2, diffs);
    diffs -= dx;
    good += dx;
    k--;
  }
  cout<<good<<endl;

  return 0;
}