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 double eps = (1e-9);
using namespace std;
int dcmp(long double a, long double b) {
  return fabsl(a - b) <= eps ? 0 : (a > b) ? 1 : -1;
}
int getBit(int num, int idx) { return ((num >> idx) & 1) == 1; }
int setBit1(int num, int idx) { return num | (1 << idx); }
long long setBit0(long long num, int idx) { return num & ~(1ll << idx); }
long long flipBit(int num, int idx) { return num ^ (1 << idx); }
void FS() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int arr[109];
int main() {
  int n;
  cin >> n;
  int mx = 0, mi = 300;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    mx = max(mx, arr[i]);
    mi = min(mi, arr[i]);
  }
  for (int i = 0; i <= 200; i++) {
    bool ok = 1;
    for (int j = 0; j < n && ok; j++) {
      for (int k = j + 1; k < n && ok; k++) {
        int diff = abs(arr[j] - arr[k]);
        if ((i != 0 && diff % i != 0) || (diff != 0 && i == 0)) ok = 0;
      }
    }
    if (ok == 0) continue;
    if (i != 0 && ((mx - mi) / i > 2)) continue;
    if (ok) {
      cout << i << endl;
      return 0;
    }
  }
  cout << "-1\n";
}
