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
#include <bits/stdc++.h>
using namespace std;
bool check(long long n, long long pos) { return n & (1LL << pos); }
long long Set(long long n, long long pos) { return n = n | (1LL << pos); }
map<vector<long long>, long long> vec;
long long arr[103];
long long lft[103];
long long rgt[103];
int main() {
  long long i, j, k, l, m, n, o, r, q;
  long long testcase;
  long long input, flag, tag, ans;
  while (cin >> n) {
    vec.clear();
    for (i = 1; i <= n; i++) {
      scanf("%lld", &arr[i]);
      lft[i] = arr[i] >> 15;
      rgt[i] = arr[i] & ((1 << 15) - 1);
    }
    for (i = 0; i < (1 << 15); i++) {
      long long prv = 0;
      vector<long long> v;
      for (j = 1; j <= n; j++) {
        long long x = lft[j] ^ i;
        long long o = __builtin_popcount(x);
        if (j != 1) v.push_back(prv - o);
        prv = o;
      }
      vec[v] = i;
    }
    long long ans = -1;
    for (i = 0; i < (1 << 15); i++) {
      long long prv = 0;
      vector<long long> v;
      for (j = 1; j <= n; j++) {
        long long x = rgt[j] ^ i;
        long long o = __builtin_popcount(x);
        if (j != 1) v.push_back(o - prv);
        prv = o;
      }
      if (vec.find(v) != vec.end()) {
        ans = (vec[v] << 15) + i;
        break;
      }
    }
    printf("%lld\n", ans);
  }
}
