Problem: Let's call two numbers similar if their binary representations contain the same number of digits equal to 1. For example:

  * 2 and 4 are similar (binary representations are 10 and 100); 
  * 1337 and 4213 are similar (binary representations are 10100111001 and 1000001110101); 
  * 3 and 2 are not similar (binary representations are 11 and 10); 
  * 42 and 13 are similar (binary representations are 101010 and 1101). 



You are given an array of n integers a_1, a_2, ..., a_n. You may choose a non-negative integer x, and then get another array of n integers b_1, b_2, ..., b_n, where b_i = a_i ⊕ x (⊕ denotes bitwise XOR).

Is it possible to obtain an array b where all numbers are similar to each other?

Input

The first line contains one integer n (2 ≤ n ≤ 100).

The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2^{30} - 1).

Output

If it is impossible to choose x so that all elements in the resulting array are similar to each other, print one integer -1.

Otherwise, print any non-negative integer not exceeding 2^{30} - 1 that can be used as x so that all elements in the resulting array are similar.

Examples

Input


2
7 2


Output


1


Input


4
3 17 6 0


Output


5


Input


3
1 2 3


Output


-1


Input


3
43 12 12


Output


1073709057

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
long long xyp(long long x, long long y) {
  if (y == 0) return 1LL;
  if (y == 1) return x;
  if (y % 2) {
    long long p = xyp(x, y - 1);
    return (x * p) % 1000000007;
  }
  long long p = xyp(x, y / 2);
  return (p * p) % 1000000007;
}
long long inv(long long x) { return xyp(x, 1000000007 - 2); }
long long fact[10000007];
long long pr[10000007];
vector<long long> rp;
void init() {
  fact[0] = 1;
  for (long long i = 1; i < 10000007; i++) {
    fact[i] = (fact[i - 1] * i) % 1000000007;
  }
}
long long ncr(long long n, long long r) {
  if (r == 0 || n == r) return 1LL;
  if (r == 1 || ((n - r) == 1)) return n;
  long long p =
      (((fact[n] * inv(fact[r])) % 1000000007) * inv(fact[n - r])) % 1000000007;
  return p;
}
void sieve() {
  memset(pr, 1, sizeof(pr));
  pr[0] = pr[1] = 0;
  for (long long i = 2; i < 10000007; i++) {
    if (pr[i]) {
      rp.push_back(i);
      for (long long j = i * i; j < 10000007; j += i) {
        pr[j] = 0;
      }
    }
  }
}
map<vector<long long>, long long> mp;
long long first(long long x) {
  long long n = 0;
  while (x > 0) {
    if (x & 1) n++;
    x >>= 1;
  }
  return n;
}
int main() {
  long long n;
  cin >> n;
  long long a[n], b[n];
  for (long long i = 0; i < n; i++) {
    long long x;
    cin >> x;
    a[i] = x >> 15;
    b[i] = x % (1 << 15);
  }
  for (long long i = 0; i < (1 << 15); i++) {
    vector<long long> v;
    for (long long j = 0; j < n; j++) {
      v.push_back(first(a[j] ^ i));
    }
    long long sub = v[0];
    for (long long j = 0; j < n; j++) {
      v[j] = v[j] - sub;
    }
    mp[v] = i;
  }
  for (long long i = 0; i < (1 << 15); i++) {
    vector<long long> v;
    for (long long j = 0; j < n; j++) {
      v.push_back(first(b[j] ^ i));
    }
    long long sub = v[0];
    for (long long j = 0; j < n; j++) {
      v[j] = sub - v[j];
    }
    if (mp.find(v) == mp.end()) continue;
    cout << ((mp[v] << 15) + i) << "\n";
    return 0;
  }
  cout << "-1\n";
  return 0;
}