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
#include <bits/stdc++.h>
using namespace std;
const int MX = 200005;
long long n, k, a[MX];
map<int, int> ini, fin;
long long acu[MX];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  for (int i = 0; i < (n); i++) cin >> a[i + 1];
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    acu[i] = a[i] + acu[i - 1];
    if (!ini[a[i]]) ini[a[i]] = i;
    fin[a[i]] = i;
  }
  long long res = acu[n];
  for (int i = 1; i <= n; i++) {
    long long l = ini[a[i]], r = fin[a[i]];
    long long x = (l - 1) * (a[i] - 1) - acu[l - 1];
    long long y = (acu[n] - acu[r]) - (n - r) * (a[i] + 1);
    int act = r - l + 1;
    if (act >= k) res = 0;
    if (l - 1 >= k - act) res = min(res, x + max(0ll, k - act));
    if (n - r >= k - act) res = min(res, y + max(0ll, k - act));
    res = min(res, x + y + max(0ll, k - act));
  }
  cout << res << '\n';
  return 0;
}
