You are given a sequence a_1, a_2, ..., a_n consisting of n integers.

You can choose any non-negative integer D (i.e. D ≥ 0), and for each a_i you can:

  * add D (only once), i. e. perform a_i := a_i + D, or 
  * subtract D (only once), i. e. perform a_i := a_i - D, or 
  * leave the value of a_i unchanged. 



It is possible that after an operation the value a_i becomes negative.

Your goal is to choose such minimum non-negative integer D and perform changes in such a way, that all a_i are equal (i.e. a_1=a_2=...=a_n).

Print the required D or, if it is impossible to choose such value D, print -1.

For example, for array [2, 8] the value D=3 is minimum possible because you can obtain the array [5, 5] if you will add D to 2 and subtract D from 8. And for array [1, 4, 7, 7] the value D=3 is also minimum possible. You can add it to 1 and subtract it from 7 and obtain the array [4, 4, 4, 4].

Input

The first line of the input contains one integer n (1 ≤ n ≤ 100) — the number of elements in a.

The second line of the input contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 100) — the sequence a.

Output

Print one integer — the minimum non-negative integer value D such that if you add this value to some a_i, subtract this value from some a_i and leave some a_i without changes, all obtained values become equal.

If it is impossible to choose such value D, print -1.

Examples

Input


6
1 4 4 7 4 1


Output


3


Input


5
2 2 5 2 5


Output


3


Input


4
1 3 3 7


Output


-1


Input


2
2 8


Output


3
#include <bits/stdc++.h>
using namespace std;
int main() {
  int t = 1;
  while (t--) {
    long long int i, y, h, j, k, r = 0, l = 2, n, m, x = 0, o = 0,
                                 cnt1 = INT_MAX, cnt2 = 0, cnt3 = 0, cnt4 = 0;
    cin >> n;
    long long int a[n], c[3004] = {0};
    set<long long int> b;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      b.insert(a[i]);
    }
    if (b.size() == 1) {
      cout << 0 << endl;
      return 0;
    }
    if (b.size() == 2) {
      set<long long int>::iterator it;
      i = 0;
      for (it = b.begin(); it != b.end(); it++) {
        c[i++] = *it;
      }
      if ((c[1] - c[0]) % 2 == 0)
        cout << (c[1] - c[0]) / 2;
      else
        cout << c[1] - c[0];
      return 0;
    }
    if (b.size() > 3) {
      cout << "-1";
    } else {
      set<long long int>::iterator it;
      i = 0;
      for (it = b.begin(); it != b.end(); it++) {
        c[i++] = *it;
      }
      x = c[1] - c[0];
      for (j = 2; j < i; j++) {
        l = c[j] - c[j - 1];
        if (l != x)
          break;
        else
          x = l;
      }
      if (j == i)
        cout << x;
      else
        cout << "-1";
    }
  }
}
