#include <bits/stdc++.h>
using namespace std;
const int N = 5002;
int n, a[N], k, b[1000002], dd[1000001];
int main() {
  ios_base::sync_with_stdio(0);
  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++) {
      int x = a[j] - a[i];
      b[x]++;
    }
  for (int i = n - k;; i++) {
    int cur = 0;
    for (int j = i; j <= 1000000; j += i) {
      cur = cur + b[j];
    }
    if (cur * 2 > (k) * (k + 1)) continue;
    int res = 0;
    for (int j(1); j <= n; j++) {
      if (dd[a[j] % i] != i)
        dd[a[j] % i] = i;
      else
        res++;
    }
    if (res <= k) {
      cout << i;
      break;
    }
  }
}

### 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;
const int N = 5002;
int n, a[N], k, b[1000002], dd[1000001];
int main() {
  ios_base::sync_with_stdio(0);
  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++) {
      int x = a[j] - a[i];
      b[x]++;
    }
  for (int i = n - k;; i++) {
    int cur = 0;
    for (int j = i; j <= 1000000; j += i) {
      cur = cur + b[j];
    }
    if (cur * 2 > (k) * (k + 1)) continue;
    int res = 0;
    for (int j(1); j <= n; j++) {
      if (dd[a[j] % i] != i)
        dd[a[j] % i] = i;
      else
        res++;
    }
    if (res <= k) {
      cout << i;
      break;
    }
  }
}

```