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.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class co_growing_sequence {

	public static int solve(int val,int prev) {
		
		String ps = Integer.toBinaryString(prev);
		String vs = Integer.toBinaryString(val);
		
		int mx = Math.max(ps.length(), vs.length());
		int mi = Math.min(ps.length(), vs.length());
		
		String z = "";
		for(int x=0;x<mx-mi;x++) {
			z = z+"0";
		}
		
		if(ps.length()==mi) {
			ps = z+ps;
		}
		if(vs.length()==mi) {
			vs = z+vs;
		}
		
		String ans = "";
		
		for(int x=0;x<mx;x++) {
			if(ps.charAt(x)=='1' && vs.charAt(x)=='0') {
				ans += "1";
			}
			else {
				ans += Character.toString(vs.charAt(x));
			}
		}
		
		return Integer.parseInt(ans,2);
	}
	
	public static void main(String[] args) throws IOException {
		// TODO Auto-generated method stub

		R_cgs.init(System.in);
		PrintWriter pw = new PrintWriter(System.out);
		
		int t = R_cgs.nextInt();
		
		for(int i=0;i<t;i++) {
			int n = R_cgs.nextInt();
			
			int[] x = new int[n];
			
			for(int a=0;a<n;a++) {
				x[a] = R_cgs.nextInt();
			}
			
			int prev = x[0];
			
			pw.print(0+" ");
			
			for(int a=1;a<n;a++) {
				prev = solve(x[a],prev);
				pw.print((prev^x[a])+" ");
			}
			
			pw.println();
		}
		
		pw.close();
	}

}

/** Class for buffered reading int and double values */

class R_cgs {
	static BufferedReader reader;
	static StringTokenizer tokenizer;

	/** call this method to initialize reader for InputStream */
	static void init(InputStream input) {
		reader = new BufferedReader(new InputStreamReader(input));
		tokenizer = new StringTokenizer("");
	}

	/** get next word */
	static String next() throws IOException {
		while (!tokenizer.hasMoreTokens()) {
			// TODO add check for eof if necessary
			tokenizer = new StringTokenizer(reader.readLine());
		}
		return tokenizer.nextToken();
	}

	static int nextInt() throws IOException {
		return Integer.parseInt(next());
	}

	static double nextDouble() throws IOException {
		return Double.parseDouble(next());
	}
	
	static long nextLong() throws IOException {
		return Long.parseLong(next());
	}
}
