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;
const long long LL_INF = (long long)2e18 + 5;
int main() {
  int n, k;
  cin >> n >> k;
  vector<long long int> a;
  a.resize(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  vector<long long int> pre(n + 1, 0), suf(n + 1, 0);
  for (int i = 0; i < n; i++) pre[i + 1] = pre[i] + a[i];
  for (int i = n - 1; i >= 0; i--) suf[i] = suf[i + 1] + a[i];
  long long int ans = LL_INF;
  for (int i = 0, j = 0; i < n; i = j) {
    long long int val = a[i];
    while (j < n && a[j] == val) j++;
    if (j - i >= k) {
      ans = 0;
      break;
    }
    int need = k - (j - i);
    int before = i;
    int after = n - j;
    long long int bc = before * val - pre[i];
    long long int ac = suf[j] - after * val;
    if (before >= need) {
      ans = min(ans, bc - (before - need));
    }
    if (after >= need) {
      ans = min(ans, ac - (after - need));
    }
    ans = min(ans, ac + bc - (before + after - need));
  }
  cout << ans << endl;
}