#include <bits/stdc++.h>
using namespace std;
int n, i, j, k, l, m, a[5005], b[1000005];
int main() {
  cin >> n >> k;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++) b[abs(a[i] - a[j])]++;
  vector<int> v(1000005, 0);
  for (int w = 1; w <= 1000005; w++) {
    int s = 0;
    for (int i = w; i <= 1000005; i += w) s += b[i];
    if (s > k * (k + 1) / 2) continue;
    int p = 0;
    for (int j = 1; j <= n; j++) {
      v[a[j] % w]++;
      if (v[a[j] % w] > 1) p++;
    }
    if (p <= k) {
      cout << w;
      return 0;
    }
    for (int j = 1; j <= n; j++) v[a[j] % w] = 0;
  }
}

### Prompt

Please formulate 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, i, j, k, l, m, a[5005], b[1000005];
int main() {
  cin >> n >> k;
  for (i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++) b[abs(a[i] - a[j])]++;
  vector<int> v(1000005, 0);
  for (int w = 1; w <= 1000005; w++) {
    int s = 0;
    for (int i = w; i <= 1000005; i += w) s += b[i];
    if (s > k * (k + 1) / 2) continue;
    int p = 0;
    for (int j = 1; j <= n; j++) {
      v[a[j] % w]++;
      if (v[a[j] % w] > 1) p++;
    }
    if (p <= k) {
      cout << w;
      return 0;
    }
    for (int j = 1; j <= n; j++) v[a[j] % w] = 0;
  }
}

```