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;
int _toggle(int N, int pos) { return N = N ^ (1 << pos); }
int _set(int N, int pos) { return N = N | (1 << pos); }
int _reset(int N, int pos) { return N = N & ~(1 << pos); }
bool _check(int N, int pos) { return (bool)(N & (1 << pos)); }
bool _upper(char a) { return a >= 'A' && a <= 'Z'; }
bool _lower(char a) { return a >= 'a' && a <= 'z'; }
bool _digit(char a) { return a >= '0' && a <= '9'; }
int dx[] = {1, -1, 0, 0, -1, -1, 1, 1};
int dy[] = {0, 0, 1, -1, -1, 1, -1, 1};
vector<int> v[2 * 100010];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  int mx = 0;
  for (int i = 1; i <= n; i++) {
    int cnt = 0, a;
    cin >> a;
    mx = max(a, mx);
    while (a) v[a].push_back(cnt++), a /= 2;
    v[a].push_back(cnt);
  }
  int ans = INT_MAX;
  for (int i = 0; i <= mx; i++) {
    int x = v[i].size();
    if (x < k) continue;
    sort(v[i].begin(), v[i].end());
    int cnt = 0;
    for (int j = 0; j < k; j++) cnt += v[i][j];
    ans = min(ans, cnt);
  }
  cout << ans << '\n';
}
