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;
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
long long powmod(long long a, long long b, long long MOD) {
  long long ans = 1;
  while (b) {
    if (b % 2) ans = ans * a % MOD;
    a = a * a % MOD;
    b /= 2;
  }
  return ans;
}
const int N = 2e5 + 11;
int n, a[N];
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  int res = 100000000;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= 1000; i++) {
    int ans = 0, q = 0, sta = 0;
    for (int j = 1; j <= n; j++) {
      if (a[j] != i && !sta) {
        sta = 1;
        ans = abs(a[j] - i);
      } else if (a[j] != i && sta) {
        if (abs(a[j] - i) != ans) {
          q = 1;
          break;
        }
      }
    }
    if (!q) res = min(res, ans);
  }
  if (res == 100000000)
    cout << -1;
  else
    cout << res << endl;
  return 0;
}
