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 JAVA):

Solution:
/*
 ID: tommatt1
 LANG: JAVA
 TASK: 
*/
import java.util.*;
import java.io.*;
public class cf1257f{

	public static void main(String[] args)throws IOException {
		PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
		BufferedReader bf=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer(bf.readLine());
		int n=Integer.parseInt(st.nextToken());
		st=new StringTokenizer(bf.readLine());
		int[] a=new int[n];
		for(int i=0;i<n;i++) {
			a[i]=Integer.parseInt(st.nextToken());
		}
		HashMap<String,Integer> map=new HashMap<>();
		int bb=1<<15;
		for(int b=0;b<bb;b++) {
			int[] key=new int[n];
			int sub=Integer.bitCount(a[0]>>15^b);
			for(int i=1;i<n;i++) {
				key[i]=Integer.bitCount(a[i]>>15^b)-sub;
			}
			String s=Arrays.toString(key);
			map.put(s, b);
		}
		for(int b=0;b<bb;b++) {
			int[] key=new int[n];
			int sub=Integer.bitCount((a[0]^b)%bb);
			for(int i=1;i<n;i++) {
				key[i]=sub-Integer.bitCount((a[i]^b)%bb);
			}
			String s=Arrays.toString(key);
			if(map.containsKey(s)) {
				out.println((map.get(s)<<15)+b);
				out.close();
				return;
			}
		}
		out.println(-1);
		out.close();
	}

}