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 N = 2e5 + 10, inf = 1e18 + 10, M = 1e5 + 10;
long long a[N], p[N], second[N], p1[N], s1[N];
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  for (long long i = 0; i < n; ++i) cin >> a[i];
  sort(a, a + n);
  long long cnt1 = 0, last1 = a[0];
  for (long long i = 0; i <= n; ++i) {
    if (a[i] == last1)
      ++cnt1;
    else {
      if (cnt1 >= k) {
        cout << 0;
        return 0;
      }
      cnt1 = 1;
    }
    last1 = a[i];
  }
  long long ans = inf, last = a[0], have = 0, cnt = 0;
  for (long long i = 0; i < n; ++i) {
    have += (a[i] - last) * i;
    if (a[i] != last) cnt = i;
    if (i >= k - 1) ans = min(ans, have - min(cnt, (i + 1 - k)));
    last = a[i];
    p[i] = have;
    p1[i] = cnt;
  }
  last = a[n - 1], have = 0, cnt = 0;
  for (long long i = n - 1; i >= 0; --i) {
    have += (last - a[i]) * (n - i - 1);
    if (a[i] != last) cnt = (n - i - 1);
    if (n - i >= k) ans = min(ans, have - min(cnt, ((n - i) - k)));
    last = a[i];
    second[i] = have;
    s1[i] = cnt;
  }
  for (long long i = 0; i < n; ++i) {
    ans = min(ans, p[i] + second[i] - min(s1[i] + p1[i], (n - k)));
    if (i != n - 1 && a[i + 1] != a[i] && a[i + 1] != a[i] + 1)
      ans = min(ans, p[i] + second[i] + i + 1 - (n - i - 1) - (n - k));
    if (i != 0 && a[i - 1] != a[i] && a[i - 1] != a[i] - 1)
      ans = min(ans, p[i] + second[i] - i + (n - i) - (n - k));
  }
  cout << ans;
}