A sequence of non-negative integers a_1, a_2, ..., a_n is called growing if for all i from 1 to n - 1 all ones (of binary representation) in a_i are in the places of ones (of binary representation) in a_{i + 1} (in other words, a_i \:\&\: a_{i + 1} = a_i, where \& denotes [bitwise AND](http://tiny.cc/xpy9uz)). If n = 1 then the sequence is considered growing as well.

For example, the following four sequences are growing: 

  * [2, 3, 15, 175] — in binary it's [10_2, 11_2, 1111_2, 10101111_2]; 
  * [5] — in binary it's [101_2]; 
  * [1, 3, 7, 15] — in binary it's [1_2, 11_2, 111_2, 1111_2]; 
  * [0, 0, 0] — in binary it's [0_2, 0_2, 0_2]. 



The following three sequences are non-growing: 

  * [3, 4, 5] — in binary it's [11_2, 100_2, 101_2]; 
  * [5, 4, 3] — in binary it's [101_2, 100_2, 011_2]; 
  * [1, 2, 4, 8] — in binary it's [0001_2, 0010_2, 0100_2, 1000_2]. 



Consider two sequences of non-negative integers x_1, x_2, ..., x_n and y_1, y_2, ..., y_n. Let's call this pair of sequences co-growing if the sequence x_1 ⊕ y_1, x_2 ⊕ y_2, ..., x_n ⊕ y_n is growing where ⊕ denotes [bitwise XOR](http://tiny.cc/bry9uz).

You are given a sequence of integers x_1, x_2, ..., x_n. Find the lexicographically minimal sequence y_1, y_2, ..., y_n such that sequences x_i and y_i are co-growing.

The sequence a_1, a_2, ..., a_n is lexicographically smaller than the sequence b_1, b_2, ..., b_n if there exists 1 ≤ k ≤ n such that a_i = b_i for any 1 ≤ i < k but a_k < b_k.

Input

The first line contains an integer t (1 ≤ t ≤ 10^4). Then t test cases follow.

The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — length of the sequence x_i.

The second line contains n integers x_1, x_2, ..., x_n (0 ≤ x_i < 2^{30}) — elements of the sequence x_i.

It is guaranteed that the sum of n overall all test cases doesn't exceed 2 ⋅ 10^5.

Output

For each test case, print n integers y_1, y_2, ..., y_n (0 ≤ y_i < 2^{30}) — lexicographically minimal sequence such that such that it's co-growing with given sequence x_i.

Example

Input


5
4
1 3 7 15
4
1 2 4 8
5
1 2 3 4 5
4
11 13 15 1
1
0


Output


0 0 0 0 
0 1 3 7 
0 1 0 3 2 
0 2 0 14 
0 
import java.util.*;
import java.lang.*;
import java.io.*;

/* Name of the class has to be "Main" only if the class is public. */
public class Codeforces {

    static int find(int target,int x)
    {
        int res=0;
        for(int j=0;j<32;j++)
        {
            int val=1<<j;
            if((x&val)==0)  // x er bit e zero
            {
                if((target&val)==0)  //arr er bit e zero
                {
                    int bit=0;

                    res=res+(bit*(int)Math.pow(2,j));
                }
                else
                {
                    int bit=1;

                    res=res+(bit*(int)Math.pow(2,j));
                }
            }
            else
            {
                if((target&val)==0)
                {
                    int bit=0;
                    res=res+(bit*(int)Math.pow(2,j));
                }
                else
                {
                    int bit=0;

                    res=res+(bit*(int)Math.pow(2,j));
                }
            }
        }
        return res;
    }
    public static void main(String[] args) throws java.lang.Exception {
        /* your code goes here */
        BufferedReader buf = new BufferedReader(new InputStreamReader(System.in));
       int t = Integer.parseInt(buf.readLine());
        StringBuilder sb = new StringBuilder();
       for (int i = 0; i < t; i++) {
           int n=Integer.parseInt(buf.readLine());
           int x[]=new int[n];
           String st1[]=(buf.readLine()).split(" ");
           for(int j=0;j<n;j++)
               x[j]=Integer.parseInt(st1[j]);
           
               int y[]=new int[n];
               y[0]=0;
               int arr[]=new int[n];
               arr[0]=y[0]^x[0];
               for(int j=1;j<n;j++)
               {
                   y[j]=find(arr[j-1],x[j]);
                   arr[j]=y[j]^x[j];
               }
               for(int j=0;j<n;j++)
                   sb.append(y[j]+" ");

           sb.append("\n");
        }
        System.out.println(sb);
    }
}