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;
const long long INF = 1e18 + 9;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  vector<long long> pref(n, 0), suf(n, 0);
  for (long long i = 1, sm = a[0]; i < n; i++) {
    pref[i] = a[i] * i - sm;
    sm += a[i];
  }
  for (long long i = n - 2, sm = a[n - 1]; i >= 0; i--) {
    suf[i] = sm - a[i] * (n - i - 1);
    sm += a[i];
  }
  long long ans = INF;
  for (long long i = 0, j = 0; i < n; i = j) {
    for (; j < n && a[j] == a[i]; j++)
      ;
    long long prev = 0, forw = 0;
    long long base = INF;
    prev += pref[i];
    prev -= i;
    if (i >= k - j + i) base = min(base, prev);
    forw += suf[j - 1];
    forw -= (n - j);
    if (n - j >= k - j + i) base = min(base, forw);
    base = min(base, prev + forw);
    ans = min(ans, base + k - (j - i));
    if (j - i >= k) ans = 0;
  }
  cout << ans << endl;
  return 0;
}