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>
const int INF = 1e8;
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  vector<int> v;
  int n, aux;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> aux;
    v.push_back(aux);
  }
  int ans = INF;
  for (int j = 0; j < 101; j++) {
    int x = 0, d = 0;
    while (abs(v[x] - j) <= 0 && x < n) x++;
    if (x < n) d = abs(v[x] - j);
    bool flag = true;
    for (int i = 1; i < n; i++) {
      if (abs(v[i] - j) != 0 && abs(v[i] - j) != d) flag = false;
    }
    if (flag) ans = min(ans, d);
  }
  if (ans == INF)
    cout << "-1\n";
  else
    cout << ans << "\n";
  return 0;
}
