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;
const int N = 105;
const int B = 15;
const int L = (1 << B) - 1;
int n, a[N];
int pc[1 << B];
map<vector<int>, int> brute(int sh) {
  map<vector<int>, int> ans;
  for (int x = 0; x <= L; x++) {
    int last = 0;
    vector<int> vec(n - 1);
    for (int i = 0; i < n; i++) {
      int num = (a[i] >> sh) & L;
      num ^= x;
      int bits = pc[num];
      if (i) {
        vec[i - 1] = bits - last;
      }
      last = bits;
    }
    ans[vec] = x;
  }
  return ans;
}
int main() {
  for (int i = 0; i < (1 << B); i++) pc[i] = __builtin_popcount(i);
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  sort(a, a + n);
  n = int(unique(a, a + n) - a);
  auto lo = brute(0);
  auto hi = brute(B);
  for (auto [vec, x] : lo) {
    auto key = vec;
    for (int &val : key) val *= -1;
    if (hi.count(key)) {
      printf("%d\n", x + (hi[key] << B));
      return 0;
    }
  }
  printf("-1\n");
}
