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>
long long inf = (long long)1e18;
using namespace std;
int expo(int base, int exponent) {
  int ans = 1;
  while (exponent != 0) {
    if (exponent & 1) ans = (1LL * ans * base) % 1000000007;
    base = (1LL * base * base) % 1000000007;
    exponent >>= 1;
  }
  return ans % 1000000007;
}
int solve() {
  long long n, d, p;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; ++i) cin >> a[i];
  sort(a.begin(), a.end());
  if ((a[n - 1] + a[0]) % 2 == 0) {
    p = (a[n - 1] + a[0]) / 2;
    d = p - a[0];
    for (long long i = 0; i < n; ++i)
      if (a[i] > p)
        a[i] -= d;
      else if (a[i] < p)
        a[i] += d;
    for (long long i = 0; i < n - 1; ++i)
      if (a[i] != a[i + 1]) {
        cout << -1;
        return 0;
      }
    cout << d;
  } else {
    d = a[n - 1] - a[0];
    for (long long i = 0; i < n; ++i)
      if (a[i] < a[n - 1]) a[i] += d;
    for (long long i = 0; i < n - 1; ++i)
      if (a[i] != a[i + 1]) {
        cout << -1;
        return 0;
      }
    cout << d;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}
