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;
bool myfunction(int i, int j) { return (i < j); }
void solve() {
  long long n;
  cin >> n;
  set<long long> st;
  long long a;
  for (long long int i = 0; i < n; i++) {
    cin >> a;
    st.insert(a);
  }
  if (st.size() > 3) {
    cout << -1;
    return;
  } else if (st.size() == 1) {
    cout << 0;
    return;
  } else if (st.size() == 2) {
    auto it = st.begin();
    int a = *it;
    it++;
    int b = *it;
    if ((a + b) % 2 == 0) {
      cout << (b - a) / 2;
    } else
      cout << (b - a);
    return;
  } else {
    auto it = st.begin();
    int a = *it;
    it++;
    int b = *it;
    it++;
    int c = *it;
    if (c - b != b - a) {
      cout << "-1";
      return;
    } else
      cout << b - a;
    return;
  }
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int test = 1;
  while (test--) solve();
  return 0;
}
