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;
typedef long long int ll;
#define fast                          \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);
#define mod 1000000007
#define print(x) cout << 'Case #' << j << ': ' << x << endl;
#define mod9 1000000009
#define mod7 1000000007
#define INF 1e18
#define sp(y) fixed << setprecision(y)
#define vi vector<int>
#define setbits(x) __builtin_popcountll(x)
#define zerobits(x) __builtin_ctzll(x)
#define pb push_back
#define mp make_pair
#define pii pair<int, int>
#define tiii tuple<int, int, int>
#define pll pair<long long, long long>
#define sii set<int>
#define sll set<long long>
#define vii vector<int>
#define vll vector<long long>
#define vll vector<long long>
#define mii map<int, int>
#define mll map<long long, long long>
#define lob lower_bound
#define upb upper_bound
#define ret return
#define present(s, x) (s.find(x) != s.end())
#define cpresent(s, x) (find(all(s), x) != s.end())
#define ford(container, it) for (auto it = container.begin(); it != container.end(); it++)
#define fors(container, it, a, b) for (auto it = a; it != b; it++)
#define ff first
#define all(v) v.begin(), v.end()
#define ss second
#define rep(i, n) for (i = 0; i < n; i++)
#define fu(i, a, n) for (i = a; i <= n; i++)
#define fd(i, n, a) for (i = n; i >= a; i--)
#define gi(n) scanf(% d, &n)
void still_single()
{
    int n;
    cin >> n;
    vll arr(n);
    int i;
    rep(i, n) cin >> arr[i];
    bool flag = true;
    rep(i, n - 1)
    {
        if (arr[i] & arr[i + 1] != arr[i])
        {
            flag = false;
            break;
        }
    }
    vll ans(n);
    ans[0] = 0;
    // if (!flag)
    {
        // cout << 0 << ' ';

        fu(i, 1, n - 1)
        {
            ll val = ans[i - 1] ^ arr[i - 1];
            ll x = val & (~arr[i]);
            ans[i] =x;
        }
        for (auto x : ans)
            cout << x << ' ';
    }
    // else
    // {
    //     rep(i, n) cout << ans[i] << ' ';
    // }
    cout << endl;
}
int main()
{
    fast int t = 1;
    cin >> t;
    while (t--)
    {
        still_single();
    }
    return 0;
}