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<algorithm>
#include<vector>

using namespace std;
typedef long long ll;

int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		int n;cin>>n;
		int a[n];
		for(int i=0;i<n;i++)
		    cin>>a[i];
		if(n==1) 
		{
		    cout<<0<<endl;
		}
		else
		{
		  //  bool flag=0;
    // 		for(int i=0;i<n-1;i++)
    // 		{
    // 		    if(a[i]&a[i+1]!=a[i])//010 100 001
    // 		    {
    // 		        flag=1;
    // 		        break;
    // 		    }
    // 		}
    // 		if(flag==0)//flag=0全部共同增长
    // 		{
    // 		    for(int i=0;i<n;i++)
    // 		    cout<<0<<' ';
    // 		    cout<<endl;
    // 		}
    // 	    else
    	    {
    	        vector<int>ans(n);
        	    ans[0]=0;
        	    for(int i=1;i<n;i++)
        	    {
        	        ll x=a[i-1]^ans[i-1];//010,100
        	        ll y=a[i];
        	        ll sum=0,val=1;
        	        while(x)
        	        {
        	            ll a1=(x&1),b1=(y&1);
        	            if(a1!=b1)
        	            {
        	                if(a1==1&&b1==0)
        	                sum+=val;
        	            }
        	            x>>=1,y>>=1,val<<=1;
        	        }
        	        ans[i]=sum;
        	    }
        	    for(int i=0;i<n;i++)
        	        cout<<ans[i]<<' ';
    	            cout<<endl;
    	    }
		}
	}
	return 0;
}