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 M 1000000007
#define MM 998244353
#define pb push_back
#define mem0(a) memset(a,0,sizeof(a))
#define mem1(a) memset(a,-1,sizeof(a))
#define memf(a) memset(a,false,sizeof(a))
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
#define sz(a) (ll)a.size()
#define F first
#define S second
#define PI 3.1415926536
#define INF 2000000000000000000
#define endl "\n"
#define _time_ 1.0 * clock() / CLOCKS_PER_SEC
#define ll unsigned long long
#define gcd __gcd
#define int long long
#define printlist(arr) for(auto x :arr) cout<< x << " ";cout<<" \n" 
#define rep(i,j,n) for(int i=j; i<n; i++)
#define yes cout << "Yes" << endl;return
#define no cout << "No" << endl;return

int fn(int x, int v)
{
  int ans = 0;
  int i =1;
  while(x>0 && v>0)
  {
    if(x%2 == 1 && v%2 == 0)
    {
      ans = ans + i;
    } 
    x/=2;
    v/=2;
    i*=2;
  }
  while(x>0)
  {
    if(x%2 == 1)
    {
      ans = ans + i;
    }
 
    x/=2;
    i*=2;
  }
  return ans;
}
void solve()
{
   int n;
   cin >> n;
   vector<int> v(n,0);
   rep(i,0,n)
   {
    cin >> v[i];
   }
   vector<int> a(n,0);
   int x = v[0];
   rep(i,1,n)
   {
    int y = fn(x,v[i]);
    a[i] = y;
    x = (v[i]^y);
   }
   rep(i,0,n)
   {
    cout << a[i] << " ";
   }
   cout << endl;
}

// Think opposite.
// Check boundary cases.
// Read all questions.

signed main()
{
  // ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
  int t = 1;
  cin >> t;
  rep(i,1,t+1)
  {
    //cout << "Case #" << i << ": ";
    solve();
  }
}