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() {
  long long int n, k;
  cin >> n >> k;
  vector<long long int> arr(n), freq(n), presum(n), postsum(n), precnt(n),
      postcnt(n);
  map<long long int, long long int> mp;
  for (long long int i = 0; i < n; i++) {
    long long int x;
    cin >> x;
    mp[x]++;
  }
  long long int i = 0;
  n = mp.size();
  for (auto x : mp) {
    arr[i] = x.first;
    freq[i] = x.second;
    i++;
  }
  presum[0] = arr[0] * freq[0];
  precnt[0] = freq[0];
  for (i = 1; i < n; i++) {
    presum[i] = (presum[i - 1] + (arr[i] * freq[i]));
    precnt[i] = (precnt[i - 1] + freq[i]);
  }
  postsum[n - 1] = arr[n - 1] * freq[n - 1];
  postcnt[n - 1] = freq[n - 1];
  for (long long int i = n - 2; i >= 0; i--) {
    postsum[i] = postsum[i + 1] + (arr[i] * freq[i]);
    postcnt[i] = postcnt[i + 1] + freq[i];
  }
  long long int ans = 2e18;
  for (i = 0; i < n; i++) {
    long long int needed = k - freq[i];
    if (needed <= 0) {
      cout << 0 << endl;
      return 0;
    }
    long long int left = (i == 0) ? 0 : precnt[i - 1];
    long long int right = (i == n - 1) ? 0 : postcnt[i + 1];
    long long int leftsum = (i == 0) ? 0 : presum[i - 1];
    long long int rightsum = (i == n - 1) ? 0 : postsum[i + 1];
    long long int leftfcont = (left * arr[i]) - leftsum;
    long long int rightfcont = rightsum - (right * arr[i]);
    long long int leftpart = (left * (arr[i] - 1)) - leftsum;
    long long int rightpart = rightsum - (right * (arr[i] + 1));
    long long int op1, op2, op3, op4;
    op1 = op2 = op3 = op4 = 2e18;
    if (left >= needed) {
      op1 = leftpart + needed;
      ans = min(ans, op1);
      op4 = rightfcont + leftpart + (needed - right);
      ans = min(ans, op4);
    }
    if (right >= needed) {
      op2 = rightpart + needed;
      ans = min(ans, op2);
      op3 = leftfcont + rightpart + (needed - left);
      ans = min(ans, op3);
    } else {
      op4 = rightfcont + leftpart + (needed - right);
      ans = min(ans, op4);
      op3 = leftfcont + rightpart + (needed - left);
      ans = min(ans, op3);
    }
  }
  cout << ans << endl;
}