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<bits/stdc++.h>
using namespace std;
#define M 1000000007
typedef long long ll;

void dfs(vector<vector<int>>&g, int n, int s, bool* vis)
{
    vis[s] = true;
    for(auto i: g[s])
    {
        if(!vis[i])
            dfs(g,n,i,vis);
    }
}
int connected_compo(vector<vector<int>>&g, int n)
{
    bool vis[n];
    memset(vis, false, n);
    
    int ans = 0;
    for(int i = 0;i<n;i++)
    {
        if(!vis[i])
        {
            ans+=1;
            dfs(g,n,i,vis);
        }
        
    }
    return ans;
}

int getmax(int *arr, int s, int e)
{
    int x = INT_MIN;
    int ans = s;
    for(int i = s;i<=e;i++)
    {
        if(arr[i] > x)
        {
            x = arr[i];
            ans = i;
        }
    }
    
    return ans;
}

void findmax(int s, int e, unordered_map<int,int>&m, int *arr, int d)
{
    if(e < s)
        return;
    int ind = getmax(arr,s,e);
    m[ind] = d;
    findmax(s,ind-1,m,arr,d+1);
    findmax(ind+1,e,m,arr,d+1);
}

int inv(int x, int m)
{
    for(int i = 1;i<m;i++)
        if((i * x)%m == 1)
            return i;
}
bool isPower(ll x, ll y)
{
    
    int res1 = log(y) / log(x);
    double res2 = log(y) / log(x); 
 
    return (res1 == res2);
}

ll power(ll n, ll k)
{
    ll res = 1;
    while(k)
    {
        if(k == 1)
            res = (res%M * n%M)%M;
        k = k>>1;
        n = (n%M*n%M)%M;
    }
    return res;
}

bool is_pos(int &k, int mv)
{
    if(mv == 0)
    {
        k++;
        return true;
    }
    if(mv <= k)
        return true;
    return false;
}
void solve()
{
    int n;
    cin >> n;
    int X[n];
    for(int i = 0;i<n;i++)
        cin >> X[i];
    vector<int>Y;
    Y.push_back(0);
    for(int i = 1;i<n;i++)
    {
        int a = 1;
        int b = 0;
        int x = X[i-1];
        int y = X[i];
        while(a <= X[i-1])
        {
            if(a&x)
            {
                if(!(a&y))
                {
                    b+=a;
                    X[i]+=a;
                }
            }
            a = a<<1;
        }
        Y.push_back(b);
    }
    for(auto i: Y)
        cout << i << " " ;

    cout << '\n';
}
int main() {
	int t;
    cin >> t;
	while(t--)
	    solve();
	return 0;
}
