#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 10;
int n, K, diff[N], vis[N];
int main() {
  cin >> n >> K;
  vector<int> a(n, 0);
  for (int i = 0; i < (n); ++i) cin >> a[i];
  sort((a).begin(), (a).end());
  for (int i = 0; i < (n); ++i)
    for (int j = i + 1; j < (n); ++j) diff[a[j] - a[i]]++;
  for (int i = 1; i < (N); ++i) {
    int cnt = 0;
    for (int j = i; j < N; j += i) cnt += diff[j];
    if (cnt > K * (K + 1) / 2) continue;
    int err = 0;
    for (int j = 0; j < (n); ++j) {
      err += vis[a[j] % i] == i;
      vis[a[j] % i] = i;
    }
    if (err <= K) {
      printf("%d\n", i);
      break;
    }
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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;
const int N = 1e6 + 10;
int n, K, diff[N], vis[N];
int main() {
  cin >> n >> K;
  vector<int> a(n, 0);
  for (int i = 0; i < (n); ++i) cin >> a[i];
  sort((a).begin(), (a).end());
  for (int i = 0; i < (n); ++i)
    for (int j = i + 1; j < (n); ++j) diff[a[j] - a[i]]++;
  for (int i = 1; i < (N); ++i) {
    int cnt = 0;
    for (int j = i; j < N; j += i) cnt += diff[j];
    if (cnt > K * (K + 1) / 2) continue;
    int err = 0;
    for (int j = 0; j < (n); ++j) {
      err += vis[a[j] % i] == i;
      vis[a[j] % i] = i;
    }
    if (err <= K) {
      printf("%d\n", i);
      break;
    }
  }
  return 0;
}

```