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>
const long long INF = 10000000000000;
const long long mod = 1e9 + 7;
using namespace std;
void add(long long& a, long long b) {
  a += b;
  if (a >= mod) a -= mod;
}
long long mul(long long a, long long b) { return (a * b) % mod; }
long long Pow(long long a, long long b) {
  if (b == 0) return 1;
  long long c = Pow(a, b / 2);
  if (b % 2) return (a * ((c * c) % mod)) % mod;
  return (c * c) % mod;
}
long long count(long long a, long long b, long long m) {
  if (a % m == 0) return b / m - a / m + 1;
  return b / m - a / m;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  cout.precision(30);
  int n, k;
  cin >> n >> k;
  std::vector<long long> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  vector<vector<long long>> v(2e5 + 1);
  for (int i = 0; i < n; i++) {
    int nb = 0;
    while (a[i]) {
      v[a[i]].push_back(nb);
      a[i] /= 2;
      nb++;
    }
  }
  long long ans = 1e18;
  for (int i = 1; i <= 2e5; i++) {
    if ((int)v[i].size() >= k) {
      sort(v[i].begin(), v[i].end());
      long long sum = 0;
      for (int j = 0; j < k; j++) sum += v[i][j];
      ans = min(ans, sum);
    }
  }
  cout << ans << endl;
  return 0;
}