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 int maxn = 100005;
using namespace std;
set<int> s;
int main() {
  int n, a;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a);
    s.insert(a);
  }
  if (s.size() > 3)
    printf("-1\n");
  else {
    if (s.size() == 1) printf("0\n");
    if (s.size() == 2) {
      int d = *(++s.begin()) - *(s.begin());
      if (d % 2 == 0) d /= 2;
      printf("%d\n", d);
    }
    if (s.size() == 3) {
      int x[3];
      int j = 0;
      for (auto t = s.begin(); t != s.end(); t++) {
        x[j++] = *t;
      }
      if (2 * x[1] != x[0] + x[2])
        printf("-1\n");
      else
        printf("%d\n", x[2] - x[1]);
    }
  }
  return 0;
}
