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;
const long long maxn = 200, N = 60000 + 20, mod = 1e9 + 7, INF = 1e9, lg = 32;
int a[maxn], b[maxn], c[maxn];
vector<int> cprime;
map<vector<int>, int> mp;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    cprime.push_back(0);
  }
  for (int i = 0; i < (1ll << 15); ++i) {
    for (int j = 0; j < n; ++j) {
      b[j] = (a[j] >> 15);
    }
    for (int j = 0; j < n; ++j) {
      b[j] ^= i;
      c[j] = __builtin_popcount(b[j]);
      cprime[j] = (c[j] - c[0]);
    }
    mp[cprime] = i;
  }
  for (int i = 0; i < (1ll << 15); ++i) {
    for (int j = 0; j < n; ++j) {
      b[j] = (a[j] & ((1ll << 15) - 1));
    }
    for (int j = 0; j < n; ++j) {
      b[j] ^= i;
      c[j] = __builtin_popcount(b[j]);
      cprime[j] = (c[0] - c[j]);
    }
    if (mp.count(cprime)) {
      long long ans = (mp[cprime] << 15) ^ i;
      cout << ans << '\n';
      return 0;
    }
  }
  cout << -1 << '\n';
}