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;
long long __gcd(long long a, long long b) { return b ? __gcd(b, a % b) : a; }
long long pwmod(long long a, long long b) {
  if (b == 0) return 1;
  long long tmp = pwmod(a, b / 2);
  tmp = (tmp * tmp) % 998244353;
  if (b & 1) tmp = (tmp * a) % 998244353;
  return tmp;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  long long arr[n];
  for (long long i = 0; i < n; i++) cin >> arr[i];
  vector<long long> cost[200001];
  for (long long i = 0; i < n; i++) {
    long long curr = 0;
    while (arr[i]) {
      cost[arr[i]].push_back(curr);
      arr[i] /= 2;
      curr++;
    }
    cost[0].push_back(curr);
  }
  long long mn = LLONG_MAX;
  for (long long i = 0; i < 200001; i++) {
    if (cost[i].size() >= k) {
      sort(cost[i].begin(), cost[i].end());
      long long sum = 0;
      for (long long j = 0; j < k; j++) {
        sum += cost[i][j];
      }
      mn = min(mn, sum);
    }
  }
  cout << mn;
}
void preset() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
int32_t main() {
  preset();
  solve();
  return 0;
  long long t;
  cin >> t;
  long long i = 1;
  while (t--) {
    solve();
    cout << '\n';
    i++;
  }
  return 0;
}