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;
template <typename TH>
void _dbg(const char *sdbg, TH h) {
  cerr << sdbg << "=" << h << "\n";
}
template <typename TH, typename... TA>
void _dbg(const char *sdbg, TH h, TA... t) {
  while (*sdbg != ',') {
    cerr << *sdbg++;
  }
  cerr << "=" << h << ",";
  _dbg(sdbg + 1, t...);
}
template <class T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <class T>
T lcm(T a, T b) {
  return a * b / gcd(a, b);
}
vector<string> split(string s) {
  istringstream buf(s);
  istream_iterator<string> beg(buf), end;
  vector<string> list(beg, end);
  return list;
}
bool isvowel(char c) {
  string s = "aeiouAEIOU";
  if (find(s.begin(), s.end(), c) != s.end()) return true;
  return false;
}
vector<long long> sieve(long long n) {
  vector<bool> prime(n + 1, true);
  prime[0] = prime[1] = 0;
  for (int i = 2; i * i <= n; i++) {
    if (prime[i])
      for (int j = i * i; j <= n; j += i) {
        prime[j] = 0;
      }
  }
  vector<long long> p;
  for (int i = 2; i <= prime.size(); i++)
    if (prime[i]) p.push_back(i);
  return p;
}
bool isprime(int n) {
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true && (n != 1);
}
int ncr(long long n, long long r) {
  long long C[r + 1];
  memset(C, 0, sizeof(C));
  C[0] = 1;
  for (long long i = 1; i <= n; i++) {
    for (long long j = min(i, r); j > 0; j--)
      C[j] = (C[j] + C[j - 1]) % 1000000007;
  }
  return C[r];
}
long long mod_exp(long long x, long long n, long long mod) {
  if (n == 0) return 1;
  if (n == 1) return x;
  long long ans = mod_exp(x, n / 2, mod);
  ans = (ans % mod * ans % mod) % mod;
  if (n & 1) ans = (ans * x) % mod;
  return ans;
}
void solve() {
  int n;
  cin >> n;
  vector<int> v(n);
  set<int> s;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    s.insert(v[i]);
  }
  int a = *s.begin();
  int b = *s.rbegin();
  int diff = b - a;
  if (s.size() > 3) {
    cout << -1;
    return;
  }
  s.erase(a);
  s.erase(b);
  if (s.size() == 1 && (diff % 2 == 1 || *s.begin() != a + diff / 2)) {
    cout << -1;
    return;
  }
  if (s.size() == 1 || diff % 2 == 0)
    cout << diff / 2;
  else
    cout << diff;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  for (int i = 1; i <= t; i++) {
    solve();
  }
  return 0;
}
