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 
#include <bits/stdc++.h>
using namespace std;
#define endl "\n"
#define int long long
#define double long double
#define ff first
#define ss second
#define pb push_back
#define pii pair<int,int>
#define all(v) (v).begin(), (v).end()

// for debug
#ifndef ONLINE_JUDGE
#define debug(x) cerr << #x <<" "; _print(x); cerr << endl;
#else
#define debug(x)
#endif

void _print(int t) {cerr << t;}
void _print(string t) {cerr << t;}
void _print(char t) {cerr << t;}
void _print(double t) {cerr << t;}

template <class T, class V> void _print(pair <T, V> p);
template <class T> void _print(vector <T> v);
template <class T> void _print(set <T> v);
template <class T, class V> void _print(map <T, V> v);
template <class T> void _print(multiset <T> v);
template <class T, class V> void _print(pair <T, V> p) {cerr << "{"; _print(p.ff); cerr << ","; _print(p.ss); cerr << "}";}
template <class T> void _print(vector <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(set <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T> void _print(multiset <T> v) {cerr << "[ "; for (T i : v) {_print(i); cerr << " ";} cerr << "]";}
template <class T, class V> void _print(map <T, V> v) {cerr << "[ "; for (auto i : v) {_print(i); cerr << " ";} cerr << "]";}

void solve()
{
	int n;
	cin >> n;
	vector<int> v(n);
	vector<vector<int>> bits(n, vector<int>(32));
	for (int i = 0; i < n; i++)
	{
		cin >> v[i];
		for (int j = 0; j < 32; j++)
		{
			if ((v[i] >> j) & 1) bits[i][j] = 1;
		}
	}
	debug(bits);
	vector<vector<int>> needed(n, vector<int>(32));
	vector<vector<int>> ans(n, vector<int>(32));
	for (int i = 0; i < 32; i++)
	{
		needed[0][i] = bits[0][i];
	}
	for (int i = 1; i < n; i++)
	{
		for (int j = 0; j < 32; j++)
		{
			if (needed[i - 1][j]) needed[i][j] = 1;
		}
		for (int j = 0; j < 32; j++)
		{
			if (needed[i][j] == 1 and bits[i][j] == 0) ans[i][j] = 1;
		}
		for (int j = 0; j < 32; j++)
		{
			needed[i][j] = bits[i][j] xor ans[i][j];
		}
	}
	vector<int> res;
	vector<int> power(32);
	power[0] = 1;
	for (int i = 1; i < 32; i++) power[i] = power[i - 1] * 2;
	for (int i = 0; i < n; i++)
	{
		int cur = 0;
		for (int j = 0; j < 32; j++)
		{
			cur += ans[i][j] * power[j];
		}
		res.pb(cur);
	}
	debug(needed);
	for (auto x : res) cout << x << " ";
	cout << endl;
}

int32_t main()
{
	ios_base:: sync_with_stdio(false);
	cin.tie(NULL); cout.tie(NULL);
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	freopen("debug.txt", "w", stderr);
#endif
#ifdef SEIVE
	seive();
#endif
	int t = 1;
	cin >> t;
	for (int i = 1; i <= t; i++)
	{
		// cout << "Case #" << i << ": ";
		solve();
	}

	return 0;
}