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 <bits/stdc++.h>
using namespace std;
#define fast()                        \
    ios_base::sync_with_stdio(false); \
    cin.tie(NULL);                    \
    cout.tie(NULL);
#define ll long long int
#define pb push_back
#define mp make_pair
#define pf push_front
#define rep(i, a, b, k) for (ll i = a; i < b; i += k)
#define rept(i, a, b) for (auto i = a; i != b; ++i)
#define drep(i, a, b, k) for (ll i = a; i >= b; i--)
#define mem0(a) memset(a, 0, sizeof(a))
#define w(t) while (t--)
#define array(arr, n)          \
    vector<ll> arr(n);         \
    for (ll i = 0; i < n; i++) \
        cin >> arr[i];
#define ni1(t) \
    ll t;      \
    cin >> t;
#define ni2(a, b) \
    ll a, b;      \
    cin >> a >> b
#define ni3(a, b, c) \
    ll a, b, c;      \
    cin >> a >> b >> c
#define ni4(a, b, c, d) \
    ll a, b, c, d;      \
    cin >> a >> b >> c >> d
#define ni5(a, b, c, d, e) \
    ll a, b, c, d, e;      \
    cin >> a >> b >> c >> d >> e
#define ni6(a, b, c, d, e, f) \
    ll a, b, c, d, e, f;      \
    cin >> a >> b >> c >> d >> e >> f
#define ns(s) \
    string s; \
    cin >> s
#define getmax(a, b) a > b ? a : b
#define getmin(a, b) a < b ? a : b
#define print(x)            \
    for (auto ele : x)      \
        cout << ele << " "; \
    cout << endl;
#define printp(x)      \
    for (auto ele : x) \
        cout << ele.firs << '	' << ele.second << endl;
#define mod 1000000007
#define siz1 200005
#define siz2 1000005
#define khatam return 0
#define all(x) x.begin(), x.end()
typedef vector<ll> vl;
typedef vector<pair<ll, ll>> vll;
typedef unordered_map<ll, ll> ull;
typedef map<ll, ll> mll;

void solve()
{
    ni1(n);
    array(val, n);
    vl ans(n, 0);
    rep(i, 0, n - 1, 1)
    {
        if (val[i] == val[i + 1])
        {
            ans[i + 1] = 0;
            continue;
        }
        else
        {

            ll no = 0;

            int aa = 0;
            int A = val[i], B = val[i + 1];
            for (int j = 0; j < 32; j++)
            {
                if ((A & (1 << j)) and !(B & (1 << j)))
                {
                    ans[i + 1] += (1 << j);
                    val[i + 1] += (1 << j);
                }
            }
        }
    }

    print(ans);
}

int main()
{
    fast();
    ni1(t);
    w(t)
    {
        solve();
    }
    khatam;
}