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 comp(const void *a, const void *b) { return *(int *)a - *(int *)b; }
int find_D(int *Arr, int num) {
  qsort(Arr, num, sizeof(int), comp);
  int numer = 0, d[2] = {0, 0};
  for (int i = 1; i < num; i++)
    if (Arr[i] > Arr[i - 1]) {
      if (numer >= 2) return -1;
      d[numer++] = Arr[i] - Arr[i - 1];
    }
  if (numer == 1) {
    if (d[0] % 2)
      return d[0];
    else
      return d[0] / 2;
  }
  if (d[0] == d[1]) return d[0];
  return -1;
}
int main() {
  int n;
  cin >> n;
  int *Mas = new int[n];
  for (int i = 0; i < n; i++) cin >> Mas[i];
  int k = find_D(Mas, n);
  cout << k;
  delete[] Mas;
  return 0;
}
