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;
void solve()
{
 
    int n;
   cin>>n;
   int a[n];
   for(int i=0;i<n;i++)
   cin>>a[i];
   if(n==1)
   {
   	cout<<0<<endl;
   	return;
    }
    bool flag=0;
    for(int i=0;i<n-1;i++)
    {
        if((a[i]&a[i+1])!=a[i])
        {
            flag=1;
            break;
        }
    }
    if(flag==0)
    {
        for(int i=0;i<n;i++)
        cout<<0<<" ";
        cout<<"\n";
        return;
    }
 
    vector<int>ans(n);
 
    ans[0]=0;
 
    for(int i=1;i<n;i++)
    {
        ll x=a[i-1]^ans[i-1];
        ll y=a[i];
        ll sum=0;
        ll val=1;
        while(x)
        {
            ll a1=(x&1);
            ll a2=(y&1);
            if(a1!=a2)
            {
                if(a1==1 && a2==0)
                sum+=val;
            }
            x>>=1;
            y>>=1;
            val<<=1;
        }
        ans[i]=sum;
    }
    for(int i=0;i<n;i++)
    cout<<ans[i]<<" ";
    cout<<"\n";
}
 
 
int main()
{
	int t;
	cin>>t;
	while(t--)
	{
		solve();
	}
}