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.math.BigInteger;
import java.util.*;
     
/*long startTime = System.currentTimeMillis(); //获取开始时间
 * long endTime = System.currentTimeMillis(); //获取结束时间
 * System.out.println("程序运行时间：" + (endTime - startTime) + "ms"); //输出程序运行时间
 */
	
    public class Main{
    	public static void main(String[] args){
    		
    		Read in=new Read(System.in);
    		
    		
    		int t=in.nextInt();
    		while(t!=0) {
    			int n=in.nextInt();
    			
    			System.out.print(0);
    			int a=in.nextInt();
    			int i=1;
    			while(i<n) {
    				int b=in.nextInt();
    				int ans=0;
    				for(int j=0;(1<<j)<=a;j++) {
    					int c=(1<<j);
    					if((a&b&c)!=(a&c)) {
    						ans+=c;
    					}
    				}
    				System.out.print(" "+ans);
    				a=ans^b;
    				i++;
    			}
    			System.out.println();
    			
    			
     			t--;
    		}
        	     	
      	}
    	
    	

	static class Read {//自定义快读 Read
	    public BufferedReader reader;
	    public StringTokenizer tokenizer;
	    public Read(InputStream stream) {
	        reader = new BufferedReader(new InputStreamReader(stream), 32768);
	        tokenizer = null;
	    }
	    public String next() {
	        while (tokenizer == null || !tokenizer.hasMoreTokens()) {
	            try {
	                tokenizer = new StringTokenizer(reader.readLine());
	            } catch (IOException e) {
	                throw new RuntimeException(e);
	            }
	        }
	        return tokenizer.nextToken();
	    }
	    public String nextLine() {
	        String str = null;
	        try {
	            str = reader.readLine();
	        } catch (IOException e) {
	            // TODO 自动生成的 catch 块
	            e.printStackTrace();
	        }
	        return str;
	    }
	    public int nextInt() {
	        return Integer.parseInt(next());
	    }
	    public long nextLong() {
	        return Long.parseLong(next());
	    }
	    public Double nextDouble() {
	        return Double.parseDouble(next());
	    }
	    public BigInteger nextBigInteger() {
	        return new BigInteger(next());
	    }
	}

}