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.io.*;
public class B{
	public static void main(String[] args)
	{
		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int t = fs.nextInt();
		me:for(int tt=0;tt<t;tt++)
		{
			int n = fs.nextInt();
			
			int[] arr = fs.readArray(n);
			
			if(n == 1)
			{
				out.println("0");
				continue me;
			}
			
			int[] ans = new int[n];
			
			for(int i=0;i<n-1;i++)
			{
				int toPut = 0;
				for(int shift=0;shift<32;shift++)
				{
					int newNum = 1<<shift;
					
					if((arr[i]&newNum) != 0)
					{
						if((arr[i+1]&newNum) == 0)
						{
							toPut |= newNum;
							arr[i+1] |= newNum;
						}
					}
					
				}
				ans[i+1] = toPut;
			}
			
			for(int i:ans)
			{
				out.print(i+" ");
			}
			out.println();
			
		}
		out.close();
	}
	static class FastScanner {
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
	public static int[] sort(int[] arr)
	{
		List<Integer> temp = new ArrayList();
		for(int i:arr)temp.add(i);
		Collections.sort(temp);
		int start = 0;
		for(int i:temp)arr[start++]=i;
		return arr;
	}
	public static String rev(String str)
	{
		char[] arr = str.toCharArray();
		char[] ret = new char[arr.length];
		int start = arr.length-1;
		for(char i:arr)ret[start--] = i;
		String ret1 = "";
		for(char ch:ret)ret1+=ch;
		return ret1;
//		return ret;
	}
}