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>
#define _GLIBCXX_DEBUG
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)
using ll = long long;
using pii = pair<int,int>;
using vi = vector<int>;
using vs = vector<string>;
using vll = vector<long long>;
using vpii = vector<pair<int,int>>;
using vivi = vector<vector<int>>;
const int dx[4] = {1,0,-1,0};
const int dy[4] = {0,1,0,-1};



int main(void) {
  int n,k;
  cin >> n >> k;
  string s;
  cin >> s;
  int cnta = 0;
  int cntb = 0;
  rep(i,n-1){
    if(s[i]=='R' && s[i+1]=='L') cnta++;
    if(s[i]=='L' && s[i+1]=='R')cntb++;
  }
  int p = 0;
  if(s[0]=='L') p++;
  if(s[n-1]=='R') p++;
  
  int q=min(2,max(k-min(cnta,cntb),0));
  
  cout << n-(2*cnta+p)+min(k,min(cnta,cntb))*2 << endl;
  
}