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.ArrayList;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Collections;
import java.util.StringTokenizer;
import java.util.Arrays;
import java.util.Comparator;
import java.util.PriorityQueue;
import java.util.List;

import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.sqrt;
import static java.lang.Math.abs;
import static java.lang.Math.random;
import static java.lang.Integer.MAX_VALUE;
import static java.lang.Integer.MIN_VALUE;
import static java.util.Collections.reverseOrder;

public final class Test {
	static int mod = 1000000007;
	static int mod1 = 998244353;
	public static void main(String[] args) {
		solve();
	}
	static long gcd(int a, int b)              {if (b == 0) return a;	return gcd(b, a % b); }
	static long modAdd(long a, long b, long m) {a = a % m; b = b % m; return (((a + b) % m) + m) % m;}
	static long modMul(long a, long b, long m) {a = a % m; b = b % m; return (((a * b) % m) + m) % m;}
	static long modSub(long a, long b, long m) {a = a % m; b = b % m; return (((a - b) % m) + m) % m;}
	static long modDiv(long a, long b, long m) {a = a % m; b = b % m; return (modMul(a, mminvprime(b, m), m) + m) % m;}  //only for prime m
	static long mminvprime(long a, long b)     {return expo(a, b - 2, b);}
	static long expo(long a, long b, long mod) {long res = 1; while (b > 0) {if ((b & 1) != 0)res = (res * a) % mod; a = (a * a) % mod; b = b >> 1;} return res;}
	static long setBits(int n)                 {int cnt = 0;  while (n != 0)   { cnt++; n = n & (n - 1);  }   return cnt;}
	static int[] sieve(int n)                  {int[] arr = new int[n + 1];  for (int i = 2; i <= n; i++)if (arr[i] == 0) {for (int j = 2 * i; j <= n; j += i)arr[j] = 1;} return arr;}
	static void debug(int[] nums)              { for (int i = 0; i < nums.length; i++)	System.out.println(nums[i] + " "); }
	static void debug(long[] nums)             { for (int i = 0; i < nums.length; i++)	System.out.println(nums[i] + " "); }
	static void debug(String[] nums)           { for (int i = 0; i < nums.length; i++)	System.out.println(nums[i] + " "); }
	static void reverse(int[] nums)            { int start = 0, end = nums.length - 1;      while (start < end) {int temp = nums[start]; nums[start] = nums[end]; nums[end] = temp; start++; end--; }  }
	static void reverse(char[] nums)            { int start = 0, end = nums.length - 1;      while (start < end) {char temp = nums[start]; nums[start] = nums[end]; nums[end] = temp; start++; end--; }  }

	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());
		}
	}

	static void sort(int[] a) {
		ArrayList<Integer> l = new ArrayList<>();
		for (int i : a) l.add(i);
		Collections.sort(l);
		for (int i = 0; i < a.length; i++) a[i] = l.get(i);
	}





//	<------------------------- CODE START FROM HERE -------------------------------->


	public static void solve() {
		FastScanner sc = new FastScanner();
		int t = sc.nextInt();
		StringBuilder sb1 = new StringBuilder();
		while (t-- != 0) {
			int n = sc.nextInt();
			int arr[] = sc.readArray(n);

			long ans[] = new long[n];

			for (int i = 1; i < n; i++) {

				long xor = (arr[i - 1] ^ ans[i - 1]);
				if ((xor & arr[i]) == xor)
					ans[i] = 0;
				else
					ans[i] = fun(xor, arr[i]);
			}
			for (long x : ans)
				sb1.append(x + " ");

			sb1.append("\n");

		}
		System.out.println(sb1);
	}
	static long fun(long a, long b) {

		long res = 0;

		for (int i = 31; i >= 0; i--) {

			if ((a & (1 << i)) != 0) {
				if ((b & (1 << i)) == 0)
					res = (res | (1 << i));
			}
		}
		return res;



	}


// <----------------------------------CODE END HERE------------------------>

// System.out.println();
// int b[] = a.clone();


}