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<iostream>
#include<queue>
#include<stack>
#include<algorithm>
#include<vector>
#include<cmath>
#include<deque>
#include<memory.h>
#include<set>
#include<map>
#include<unordered_map>
#include<unordered_set>
#include<cstdio>
#include<cstring>
#include<string.h>
#include<algorithm>
#include<string>
#include<fstream>
#include<math.h>
using namespace std;
#define ll long long
#define ull unsigned long long
#define Design ios::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define db double 
#define fi first
#define sc second
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pb push_back
#define vi vector<int>
#define vl vector<ll>
#define minf 0x3f3f3f3f3f3f3f3f
ll GCD(ll a, ll b) { return b ? GCD(b, a % b) : a; }
ll LCM(ll a, ll b) { return a * b / GCD(a, b); }
ll lb(ll x) { return x & (-x); }
const ll N = 2e5 + 7;
const ll inf = 1e18 + 1;
const ll mod2 = 998244353;
const ll mod1 = 1e9 + 7;
const int P1 = 13331;
const int P2 = 131;
const double eps = 1e-6;
const db dbinf = 1e13 + 7;
const ll mod3 = 1e6 + 3;
const db pi=acos(-1);
int t;
ll n,a[N],b[N],A[40],B[40];
void fill(ll* a,ll x){
	for(int i=30;i>=0;i--){
		if(x&1)a[i]=1;
		else a[i]=0;
		x>>=1;
	}
}
int main() {
	cin>>t;
	while(t--){
		cin>>n;
		for(int i=1;i<=n;i++)cin>>a[i];
		b[1]=0;
		for(int i=2;i<=n;i++){
			fill(A,a[i-1]^b[i-1]);
			fill(B,a[i]);
			ll now=0;
			for(int i=0;i<=30;i++){
				if(A[i]){
					if(B[i]){
						now<<=1;
					}else{
						now<<=1;
						now^=1;
					}
				}else{
					if(A[i]){
						now<<=1;
						now^=1;
					}else{
						now<<=1;
					}
				}
			}
			b[i]=now;
		}
		for(int i=1;i<=n;i++)cout<<b[i]<<" ";cout<<"\n";
	}
	return 0;
}