#include <bits/stdc++.h>
using namespace std;
int s[5010];
int n, k;
short conf[1000010][5];
int len[1000010];
int mark[5010];
void solve() {
  sort(s, s + n);
  for (int i = 0; i < n; ++i) {
    for (int u = i + 1; u < n; ++u) {
      const int diff = s[u] - s[i];
      if (len[diff] < 5) conf[diff][len[diff]++] = u;
    }
  }
  for (int put = 1;; ++put) {
    int remove = 0;
    for (int u = put; remove <= k && u < 1000001; u += put) {
      for (int i = 0; i < len[u]; ++i) {
        remove += mark[conf[u][i]] != put;
        mark[conf[u][i]] = put;
      }
    }
    if (remove <= k) {
      cout << put << endl;
      break;
    }
  }
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) cin >> s[i];
  solve();
  return 0;
}

### Prompt

Please create a solution in Cpp 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 s[5010];
int n, k;
short conf[1000010][5];
int len[1000010];
int mark[5010];
void solve() {
  sort(s, s + n);
  for (int i = 0; i < n; ++i) {
    for (int u = i + 1; u < n; ++u) {
      const int diff = s[u] - s[i];
      if (len[diff] < 5) conf[diff][len[diff]++] = u;
    }
  }
  for (int put = 1;; ++put) {
    int remove = 0;
    for (int u = put; remove <= k && u < 1000001; u += put) {
      for (int i = 0; i < len[u]; ++i) {
        remove += mark[conf[u][i]] != put;
        mark[conf[u][i]] = put;
      }
    }
    if (remove <= k) {
      cout << put << endl;
      break;
    }
  }
}
int main() {
  cin >> n >> k;
  for (int i = 0; i < n; ++i) cin >> s[i];
  solve();
  return 0;
}

```