#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define repa(i,a,b) for(ll i = a; i <= b; i++)
#define repd(i,a,b) for(ll i = a; i >= b; i--)
#define pb push_back
#define ff first
#define ss second
#define vi vector<int>
#define vl vector<ll>
#define IOS ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define ld long double
#define nl cout << "\n"
#define MAX 1000000000000
#define all(x) x.begin(),x.end()
//debuging	
#define deb1(n) cout<<n<<" "
#define deb2(a,n) cout<<"a->"<<n<<" "
#define deb3(A,a,b) repa(i,a,b){cout<<A[i]<<" ";}nl
#define deb4(v) repa(i,0,v.size()-1){cout<<v[i]<<" ";}nl
// #define printclock cerr<<"Time : "<<1000*(ld)clock()/(ld)CLOCKS_PER_SEC<<"ms\n";

void solve(){
	ll n;
	cin >> n;
	vl v(n), ans(n,0);
	repa(i,0,n-1) cin >> v[i];
	if(n == 1){
		cout << 0; nl;
		return;
	}
	ans[0] = 0;
	bitset <32> bst1, bst2;
	repa(i,1,n-1){
		bst1 = v[i-1] ^ ans[i-1];
		bst2 = v[i];
		for(ll j = 0; j <= 30 ; j++){
			if( bst1[j] == 1 && bst2[j] == 0){
				ans[i] += 1<<j;
			}
		}

	}
	repa(i,0,n-1){
		cout << ans[i] << " ";
	}
	nl;
}

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

	ll t = 1;
	cin >> t;
	while(t--){
		solve();
	}
  	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define repa(i,a,b) for(ll i = a; i <= b; i++)
#define repd(i,a,b) for(ll i = a; i >= b; i--)
#define pb push_back
#define ff first
#define ss second
#define vi vector<int>
#define vl vector<ll>
#define IOS ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define ld long double
#define nl cout << "\n"
#define MAX 1000000000000
#define all(x) x.begin(),x.end()
//debuging	
#define deb1(n) cout<<n<<" "
#define deb2(a,n) cout<<"a->"<<n<<" "
#define deb3(A,a,b) repa(i,a,b){cout<<A[i]<<" ";}nl
#define deb4(v) repa(i,0,v.size()-1){cout<<v[i]<<" ";}nl
// #define printclock cerr<<"Time : "<<1000*(ld)clock()/(ld)CLOCKS_PER_SEC<<"ms\n";

void solve(){
	ll n;
	cin >> n;
	vl v(n), ans(n,0);
	repa(i,0,n-1) cin >> v[i];
	if(n == 1){
		cout << 0; nl;
		return;
	}
	ans[0] = 0;
	bitset <32> bst1, bst2;
	repa(i,1,n-1){
		bst1 = v[i-1] ^ ans[i-1];
		bst2 = v[i];
		for(ll j = 0; j <= 30 ; j++){
			if( bst1[j] == 1 && bst2[j] == 0){
				ans[i] += 1<<j;
			}
		}

	}
	repa(i,0,n-1){
		cout << ans[i] << " ";
	}
	nl;
}

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

	ll t = 1;
	cin >> t;
	while(t--){
		solve();
	}
  	return 0;
}
```