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;
inline int nextint() {
  int x;
  scanf("%d", &x);
  return x;
}
inline long long nextll() {
  long long x;
  scanf("%lld", &x);
  return x;
}
template <typename T>
void scanint(T &x) {
  T c = getchar();
  while (((c < 48) || (c > 57)) && (c != '-')) c = getchar();
  bool neg = false;
  if (c == '-') neg = true;
  x = 0;
  for (; c < 48 || c > 57; c = getchar())
    ;
  for (; c > 47 && c < 58; c = getchar()) x = (x * 10) + (c - 48);
  if (neg) x = -x;
}
template <typename T>
T min_(T a, T b) {
  return a > b ? b : a;
}
template <typename T, typename... Ts>
T min_(T first, Ts... last) {
  return min_(first, min_(last...));
}
template <class T, class S>
std::ostream &operator<<(std::ostream &os, const std::pair<T, S> &t) {
  os << "(" << t.first << ", " << t.second << ")";
  return os;
}
template <typename T>
ostream &operator<<(ostream &out, const vector<T> &v) {
  out << "[";
  size_t last = v.size() - 1;
  for (size_t i = 0; i < v.size(); ++i) {
    out << v[i];
    if (i != last) out << ", ";
  }
  out << "]";
  return out;
}
long long pwr(long long base, long long p, long long mod) {
  long long ans = 1;
  while (p) {
    if (p & 1) ans = (ans * base) % mod;
    base = (base * base) % mod;
    p /= 2;
  }
  return ans;
}
long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * (b / gcd(a, b)); }
const long double PI = (long double)(3.1415926535897932384626433832795);
const long long mx_ll = numeric_limits<long long>::max();
const int mx_int = numeric_limits<int>::max();
const int mod = 1e9 + 7;
int cnt[1001];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  vector<int> inp;
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    inp.push_back(x);
    cnt[x] += 1;
  }
  if (n == 1 || (cnt[inp[0]] == n)) {
    cout << 0 << "\n";
    return 0;
  }
  sort(inp.begin(), inp.end());
  int D1 = inp[n - 1] - inp[0];
  bool ok1 = true;
  for (int i = 0; i <= n - 2; i++) {
    if (inp[i] == inp[n - 1]) continue;
    if (inp[i] + D1 != inp[n - 1]) {
      ok1 = false;
      break;
    }
  }
  int D2 = (D1 % 2 == 0) ? D1 / 2 : -1;
  if (D2 == -1) {
    if (ok1) {
      cout << D1 << "\n";
    } else {
      cout << -1 << "\n";
    }
  } else {
    if (cnt[inp[0]] + cnt[D2 + inp[0]] + cnt[inp[n - 1]] != n) {
      cout << -1 << "\n";
    } else {
      cout << D2 << "\n";
    }
  }
  return 0;
}
