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 
// Author : shreyas
// McMurdo Station, Antartica
// Pengiun Intelligence Agency
// Date :

#include <bits/stdc++.h>
using namespace std;

#define fast                    ios_base::sync_with_stdio(false);  cin.tie(NULL);
#define time                    cerr<<"Time taken : " <<(float)clock()/CLOCKS_PER_SEC <<" secs"<<"\n"  ;
#define F                       first
#define S                       second
#define pb                      push_back
typedef long long int           ll  ;


const ll INF = (ll)1e18   ;
const ll MOD = (ll)1e9 + 7 ;

#define TRACE
#ifdef TRACE
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
	cerr << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
	const char* comma = strchr(names + 1, ','); cerr.write(names, comma - names) << " : " << arg1 << " | "; __f(comma + 1, args...);
}
#else
#define trace(...)
#endif



ll GCD (ll a , ll b) {
	if (b == 0) {
		return a  ;
	}

	return  GCD(b , a % b) ;
}

ll LCM(ll a , ll b) {
	return (a * b) / GCD(a , b)   ;
}

vector<bool>sieveOfErato(ll n ) {

	vector<bool>isPrime(n + 1 , true ) ;

	isPrime[1] = false ;
	isPrime[0] = false ;

	for (ll i = 2 ; i * i <= n; i++) {

		for (ll j = 2 * i ; j <= n; j += i) {
			isPrime[j] = false;
		}
	}

	return isPrime ;
}


/*-------------------------------------------------------------------------------------------------------------*/

void solve() {

	int n ;
	cin >> n;

	int a[n]  ;

	for (int i = 0 ; i < n ; i++)
		cin >> a[i]  ;

	bool ok = 0 ;

	for (int i =  1; i < n ; i++) {
		if (a[i] & a[i - 1] != a[i - 1]) {
			ok = 1;
			break ;
		}
	}

	if (ok) {
		for (int i = 0 ; i < n ; i++)
			cout << 0 << " " ;
	}
	else {

		vector<int>ans ;
		ans.pb(0)  ;

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

			int r  = (a[i - 1] ^ a[i])  ;

			r = r & a[i - 1] ;

			ans.pb(r)  ;

			a[i] = a[i] ^ r ;
		}

		for (auto j : ans) {
			cout << j << " " ;
		}
	}

	cout << "\n" ;

}


int32_t main() {

	fast ; time;

#ifndef ONLINE_JUDGE
	freopen("input.txt" , "r" , stdin)  ;
	freopen("output.txt" , "w" , stdout)  ;
#endif

	int t = 1;
	cin >> t;

	while (t--) {
		solve()  ;
	}


	return 0  ;
}
