Problem: You are given the array a consisting of n elements and the integer k ≤ n.

You want to obtain at least k equal elements in the array a. In one move, you can make one of the following two operations:

  * Take one of the minimum elements of the array and increase its value by one (more formally, if the minimum value of a is mn then you choose such index i that a_i = mn and set a_i := a_i + 1); 
  * take one of the maximum elements of the array and decrease its value by one (more formally, if the maximum value of a is mx then you choose such index i that a_i = mx and set a_i := a_i - 1). 



Your task is to calculate the minimum number of moves required to obtain at least k equal elements in the array.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5) — the number of elements in a and the required number of equal elements.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9), where a_i is the i-th element of a.

Output

Print one integer — the minimum number of moves required to obtain at least k equal elements in the array.

Examples

Input


6 5
1 2 2 4 2 3


Output


3


Input


7 5
3 3 2 1 1 1 3


Output


4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  map<long long, long long> mm;
  long long arr[n];
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    mm[x]++;
    arr[i] = x;
  }
  long long cur = 0;
  sort(arr, arr + n);
  for (long long i = 0; i < n; i++) {
    long long j = i;
    while (arr[j] == arr[i] && j < n) j++;
    cur = max(cur, j - i);
    i = j - 1;
  }
  cerr << cur << " " << k;
  for (long long i = 0; i < n; i++) cerr << " " << arr[i];
  if (cur >= k)
    cout << "0";
  else {
    map<long long, long long> front;
    map<long long, long long> back;
    long long A = 0, B = 0;
    auto itr = mm.begin();
    long long mn = itr->first;
    long long freq = itr->second;
    long long cost = 0;
    itr++;
    for (; itr != mm.end(); itr++) {
      long long val = itr->first;
      long long FR = itr->second;
      cost += (val - mn) * freq;
      freq = FR + freq;
      mn = val;
      front[val] = cost;
      if (freq >= k && A == 0) A = cost - (freq - k);
    }
    auto i = mm.rbegin();
    long long mx = i->first;
    freq = i->second;
    i++;
    long long costt = 0;
    for (; i != mm.rend(); ++i) {
      long long val = i->first;
      long long FR = i->second;
      costt += (mx - val) * freq;
      freq = FR + freq;
      mx = val;
      back[val] = costt;
      if (freq >= k && B == 0) B = costt - (freq - k);
    }
    long long ans = min(A, B);
    for (long long i = 1; i < n - 1; i++) {
      long long val = arr[i];
      ans = min(ans, front[val] + back[val] - (n - k));
    }
    cout << ans;
  }
  return 0;
}