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.io.*;
import java.util.*;

public class Main {

  static class Pair implements Comparable{
    int f;
    int s;
    int[] l;
    Pair(int f, int s){
      this.f = f;
      this.s = s;
    }
    Pair(int[] l){
      this.l = l;
    }
    Pair(int f, int s, int flag){
      if(f > s){
        this.f = s;
        this.s = f;
      }
      else{
        this.f = f;
        this.s = s;
      }
    }

    @Override
    public int compareTo(Object o) {
      Pair p = (Pair) o;
      if(p.f > this.f){
        return -1;
      }
      else if(p.f == this.f){
        return 0;
      }
      else{
        return 1;
      }
    }
  }

  public static void solve(Pair a, Pair b, Pair c){

  }




  public static void main(String[] args) throws IOException {
    Scanner sc = new Scanner(System.in);
    int t = sc.nextInt();
    while(t-- > 0) {
      int n = sc.nextInt();
      long[] a = new long[n];
      for(int i = 0 ; i < n ; i++){
        a[i] = sc.nextLong();
      }

      long prev = 0;
      System.out.print(prev + " ");
      for(int i = 1 ; i < n ; i++){
        long xor_prev = prev ^ a[i-1];

        long xor_curr = xor_prev | a[i];

        prev = xor_curr ^ a[i];

        System.out.print(prev + " ");
      }
      System.out.println();

    }
  }

}
