#include <bits/stdc++.h>
using namespace std;
const int N = 1000010;
int n, k;
int a[N];
int c[N], flag[N];
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
  for (int i = 0; i < n; ++i)
    for (int j = i + 1; j < n; ++j) {
      int tem = abs(a[i] - a[j]);
      c[tem]++;
    }
  for (int i = 1;; ++i) {
    int cnt = 0, res = 0;
    for (int j = i; j < N; j += i) cnt += c[j];
    if (cnt > k * (k + 1) / 2) continue;
    for (int j = 0; j < n; ++j) {
      if (flag[a[j] % i] == i)
        res++;
      else
        flag[a[j] % i] = i;
    }
    if (res <= k) {
      printf("%d\n", i);
      break;
    }
  }
  return 0;
}

### Prompt

Your challenge is to write 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 = 1000010;
int n, k;
int a[N];
int c[N], flag[N];
int main() {
  scanf("%d %d", &n, &k);
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
  for (int i = 0; i < n; ++i)
    for (int j = i + 1; j < n; ++j) {
      int tem = abs(a[i] - a[j]);
      c[tem]++;
    }
  for (int i = 1;; ++i) {
    int cnt = 0, res = 0;
    for (int j = i; j < N; j += i) cnt += c[j];
    if (cnt > k * (k + 1) / 2) continue;
    for (int j = 0; j < n; ++j) {
      if (flag[a[j] % i] == i)
        res++;
      else
        flag[a[j] % i] = i;
    }
    if (res <= k) {
      printf("%d\n", i);
      break;
    }
  }
  return 0;
}

```