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;
const int oo = 1e9 + 9;
const long long inf = 1e18 + 18;
const int max6 = 1e6 + 6;
const int modx = 1e9 + 7;
const int mody = 997;
const int base = 137;
int a[max6];
set<int> s;
int main() {
  ios_base::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  int res = oo;
  for (int i = 0; i <= 100; ++i) {
    s.clear();
    bool ok = true;
    for (int j = 1; j <= n; ++j) s.insert(abs(a[j] - i));
    if (s.size() > 2) ok = false;
    if (s.size() == 2) {
      int tmp1 = *s.begin();
      int tmp2 = *s.rbegin();
      if (tmp1 != tmp2 && tmp1 != 0) ok = false;
    }
    if (ok) res = min(res, *s.rbegin());
  }
  if (res == oo)
    cout << -1;
  else
    cout << res;
}
