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;
vector<multiset<long long> > ve;
vector<long long> flag;
int n, k;
inline void pushint(int num, int ci) {
  if (ve[num].size() == k) {
    if (*ve[num].rbegin() > ci) {
      flag[num] -= *ve[num].rbegin();
      auto temp = ve[num].end();
      temp--;
      ve[num].erase(temp);
      ve[num].insert(ci);
      flag[num] += ci;
    }
  } else {
    ve[num].insert(ci);
    flag[num] += ci;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  if (k == 1) {
    long long t;
    for (int i = 0; i < n; i++) {
      cin >> t;
    }
    cout << 0 << endl;
    return 0;
  }
  ve.resize(210000);
  flag.resize(210000);
  for (int i = 0; i < n; i++) {
    int t;
    cin >> t;
    int jici = 0;
    while (t) {
      pushint(t, jici);
      t /= 2;
      jici++;
    }
    pushint(0, jici);
  }
  long long value = 9999999999;
  for (int i = 0; i <= 200010; i++) {
    if (ve[i].size() == k && flag[i] < value) value = flag[i];
  }
  cout << value << endl;
}
