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 inf = 2e9;
const int maxn = 1e6 + 10;
int d[4][2] = {1, 0, -1, 0, 0, 1, 0, -1};
inline long long read() {
  long long s = 0, w = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (isdigit(ch)) s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
inline void write(long long x) {
  if (x < 0) putchar('-'), x = -x;
  if (x > 9) write(x / 10);
  putchar(x % 10 + '0');
}
int gcd(int x, int y) { return y == 0 ? x : gcd(y, x % y); }
long long a[105];
vector<int> v, vv;
map<vector<int>, long long> mp;
int get(int x) {
  int num = 0;
  while (x) {
    if (x & 1) {
      num++;
    }
    x = x >> 1;
  }
  return num;
}
int main() {
  int n;
  n = read();
  for (int i = 1; i <= n; i++) scanf("%lld", &(a[i]));
  for (int i = 0; i < (1LL << 15); i++) {
    v.clear();
    vv.clear();
    for (int k = 1; k <= n; k++) v.push_back(get((a[k] >> 15) ^ i));
    for (int k = 0; k <= n - 2; k++) vv.push_back(v[k] - v[k + 1]);
    mp[vv] = i;
  }
  for (int i = 0; i < (1LL << 15); i++) {
    v.clear();
    vv.clear();
    for (int k = 1; k <= n; k++) v.push_back(get((a[k] % (1 << 15)) ^ i));
    for (int k = 1; k <= n - 1; k++) vv.push_back(v[k] - v[k - 1]);
    if (mp.count(vv)) {
      write((mp[vv] << 15) + i);
      puts("");
      return 0;
    }
  }
  write(-1);
  puts("");
  return 0;
}
