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;
using ll = long long;
using ii = pair<int, int>;
using vi = vector<int>;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3fLL;
const double err = 1e-9;
const int mod = 1e9 + 7;
const int N = 1e5 + 10;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  int n;
  cin >> n;
  int a[101];
  for (int(i) = (0); (i) < (n); (i)++) cin >> a[i];
  sort(a, a + n);
  int mx = a[n - 1];
  int mi = a[0];
  bool kkk = true;
  for (int(i) = (0); (i) < (n); (i)++)
    if (a[i] != mx and a[i] != mi) kkk = false;
  int d = (mx - mi) / 2;
  if (kkk) {
    if ((mx - mi) % 2 == 0)
      cout << d << endl;
    else
      cout << mx - mi << endl;
    return 0;
  }
  int tg = mi + d;
  for (int(i) = (0); (i) < (n); (i)++)
    if (a[i] < tg)
      a[i] += d;
    else if (a[i] > tg)
      a[i] -= d;
  for (int(i) = (1); (i) < (n); (i)++) {
    if (a[i] != a[i - 1]) {
      cout << -1 << endl;
      return 0;
    }
  }
  cout << d << endl;
  return 0;
}
