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;
int leap[200], a[100], sum;
int main() {
  int n;
  while (~scanf("%d", &n)) {
    memset(leap, 0, sizeof(leap));
    sum = 0;
    memset(a, 0, sizeof(a));
    for (int i = (0); i < (n); ++i) {
      int x;
      scanf("%d", &x);
      if (leap[x] == 0) {
        leap[x]++;
        a[++sum] = x;
      }
    }
    if (sum == 1) {
      printf("0\n");
      continue;
    } else if (sum == 2) {
      int d = 0, mi = 0, ma = 0;
      mi = min(a[1], a[2]);
      ma = max(a[1], a[2]);
      d = ma + mi;
      if (d == ((d / 2) * 2))
        printf("%d\n", ma - d / 2);
      else
        printf("%d\n", ma - mi);
      continue;
    } else if (sum > 3) {
      printf("-1\n");
      continue;
    }
    sort(a + 1, a + sum + 1);
    if (2 * a[2] == a[1] + a[3])
      printf("%d\n", a[3] - a[2]);
    else
      printf("-1\n");
  }
}
