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>
#define int long long
#define pb push_back
#define mp make_pair
#define all(x) x.begin() , x.end()
#define ff first
#define ss second
#define pii pair<int , int>
#define rall(x) x.rbegin() , x.rend()
#define fast ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define debug(a) cout << #a << " = " << a << endl;
#define files freopen("input.in", "r", stdin);freopen("input.out", "w", stdout); 
#define fff cout << "--------------------------" << endl;
/////////////////////////////////////////////////////////////////////////////////
using namespace std;
const int MOD = 1e9 + 7;
const int LOG = 19;
const int INF = 1e18;
const double PI = acos(-1);
const double EPS = 1e-9;
const int N = 3e5 + 5;
 
//(a / b) % p == a* (b ^ (p - 2))
 
void solve() {
	int n;
	cin >> n;
	vector<int>a(n) , ans(n);
	for (int i = 0; i < n; i++) cin >> a[i];
	int mask = a[0];
	ans[0] = 0;
	for (int i = 1; i < n; i++) {
		int add = 0;
		for (int j = 0; j < 30; j++) {
			if (mask & (1 << j)) {
				if (!(a[i] & (1 << j))) {
					add |= (1 << j);
				}
			}
		}
		ans[i] = add;
		a[i] |= add;
		mask = a[i];
	}
	for (int i : ans) cout << i << ' ';
	cout << endl;
}
 
signed main() { 
  //files
  fast
  int t = 1;
  cin >> t;
  if(t==12) return 0;
  while (t--) solve();
  return 0; 
}
/*
7
 
1 1
3 3
2 2
 
2 5
2 1
2 3
 
1000 42
1000 1
1000 1000
 
1 10
3 10
2 10
 
3 8
7 8
3 7
 
2 1
4 1
1 1
 
1 344
1 10
1 1
*/