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>
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("avx2")
#pragma GCC optimize("Os")
using namespace std;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  string str;
  long long i, j, k, x = 0, y = 0, n, m, t, sum = 0, a = 0, b, l, c, d, id;
  cin >> n;
  set<long long> s;
  for (i = 0; i < n; i++) {
    cin >> a;
    sum += a;
    s.insert(a);
  }
  x = s.size();
  if (x >= 4)
    cout << -1;
  else {
    if (x == 1)
      cout << 0;
    else if (x == 2) {
      a = *next(s.begin(), 0);
      b = *next(s.begin(), 1);
      if ((a + b) % 2 != 0)
        cout << b - a;
      else
        cout << (b - a) / 2;
    } else {
      a = *next(s.begin(), 0);
      b = *next(s.begin(), 1);
      c = *next(s.begin(), 2);
      t = c - b;
      if (a + c == b * 2)
        cout << t;
      else
        cout << -1;
    }
  }
  return 0;
}
