#include <bits/stdc++.h>
using namespace std;
int mod(int a) { return a > 0 ? a : -a; }
int a[5000], p[1000000 + 1], vis[1000000 + 1];
int main() {
  int n, k, i, j, m, t;
  cin >> n >> k;
  for (i = 0; i < n; i++) cin >> a[i];
  for (i = 0; i < n; i++)
    for (j = i + 1; j < n; j++) p[mod(a[i] - a[j])]++;
  for (m = 1; m <= 1000000; m++) {
    for (i = 2; m * i <= 1000000; i++) p[m] += p[m * i];
    if (p[m] > k * (k + 1) / 2) continue;
    t = 0;
    for (i = 0; i < n; i++) {
      if (vis[(a[i]) % m] == m) t++;
      vis[(a[i]) % m] = m;
    }
    if (t <= k) break;
  }
  cout << m << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

You have been given n distinct integers a1, a2, ..., an. You can remove at most k of them. Find the minimum modular m (m > 0), so that for every pair of the remaining integers (ai, aj), the following unequality holds: <image>.

Input

The first line contains two integers n and k (1 ≤ n ≤ 5000, 0 ≤ k ≤ 4), which we have mentioned above. 

The second line contains n distinct integers a1, a2, ..., an (0 ≤ ai ≤ 106).

Output

Print a single positive integer — the minimum m.

Examples

Input

7 0
0 2 3 6 7 12 18


Output

13


Input

7 1
0 2 3 6 7 12 18


Output

7

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int mod(int a) { return a > 0 ? a : -a; }
int a[5000], p[1000000 + 1], vis[1000000 + 1];
int main() {
  int n, k, i, j, m, t;
  cin >> n >> k;
  for (i = 0; i < n; i++) cin >> a[i];
  for (i = 0; i < n; i++)
    for (j = i + 1; j < n; j++) p[mod(a[i] - a[j])]++;
  for (m = 1; m <= 1000000; m++) {
    for (i = 2; m * i <= 1000000; i++) p[m] += p[m * i];
    if (p[m] > k * (k + 1) / 2) continue;
    t = 0;
    for (i = 0; i < n; i++) {
      if (vis[(a[i]) % m] == m) t++;
      vis[(a[i]) % m] = m;
    }
    if (t <= k) break;
  }
  cout << m << endl;
  return 0;
}

```