#include <bits/stdc++.h>
using namespace std;
int n, k, ans;
int a[1000007], cnt[1000007], haved[1000007];
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  for (int i = 1; i < n; i++)
    for (int j = i + 1; j <= n; j++) cnt[a[j] - a[i]]++;
  for (int mod = 1;; mod++) {
    int cntt = 0, step = 0;
    for (int i = mod; i < 1000007; i += mod) cntt += cnt[i];
    if (cntt > k * (k + 1) / 2) continue;
    for (int i = 1; i <= n; i++) {
      if (haved[a[i] % mod] == mod) step++;
      haved[a[i] % mod] = mod;
    }
    if (step <= k) return cout << mod, 0;
  }
}

### Prompt

Generate a CPP solution to the following problem:

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 n, k, ans;
int a[1000007], cnt[1000007], haved[1000007];
int main() {
  cin >> n >> k;
  for (int i = 1; i <= n; i++) cin >> a[i];
  sort(a + 1, a + n + 1);
  for (int i = 1; i < n; i++)
    for (int j = i + 1; j <= n; j++) cnt[a[j] - a[i]]++;
  for (int mod = 1;; mod++) {
    int cntt = 0, step = 0;
    for (int i = mod; i < 1000007; i += mod) cntt += cnt[i];
    if (cntt > k * (k + 1) / 2) continue;
    for (int i = 1; i <= n; i++) {
      if (haved[a[i] % mod] == mod) step++;
      haved[a[i] % mod] = mod;
    }
    if (step <= k) return cout << mod, 0;
  }
}

```