#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")
#include <bits/stdc++.h>
#define fast_io  ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define pb push_back
#define ll long long
#define pb push_back
#define mp make_pair
using namespace std;

int t, n ;
vector<int> x, y ;
void solve() {
 vector<int> y(n+1, 0) ;
 vector<int> res(n+1) ;
 res[0] = x[0] ;
 y[0] = 0 ;
 
 for(int i = 1 ; i < n ; i++) {
   for(int pos = 31 ; pos >= 0 ; pos--) {
     if(res[i-1] & (1<<pos)) {
        int x2 = ((x[i] & (1<<pos)) > 0) ;
        int y2 = x2 ^ 1 ;
        if(y2 == 1) {
          y[i] = y[i] | (1<<pos) ;
        }
     }
     res[i] = x[i] ^ y[i] ;
   }
 }
 
 for(int i = 0 ; i < n ; i++) 
   cout << y[i] << " " ;
}

int main() {
  fast_io
  cin >> t ;
  while(t--) {
    cin >> n ;
    x.resize(n) ;
    for(int i = 0 ; i < n ; i++) {
      cin >> x[i] ;
    }
    solve() ;
    cout << "\n" ;
  }
return 0 ;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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 

### Response

```cpp
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")
#include <bits/stdc++.h>
#define fast_io  ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
#define pb push_back
#define ll long long
#define pb push_back
#define mp make_pair
using namespace std;

int t, n ;
vector<int> x, y ;
void solve() {
 vector<int> y(n+1, 0) ;
 vector<int> res(n+1) ;
 res[0] = x[0] ;
 y[0] = 0 ;
 
 for(int i = 1 ; i < n ; i++) {
   for(int pos = 31 ; pos >= 0 ; pos--) {
     if(res[i-1] & (1<<pos)) {
        int x2 = ((x[i] & (1<<pos)) > 0) ;
        int y2 = x2 ^ 1 ;
        if(y2 == 1) {
          y[i] = y[i] | (1<<pos) ;
        }
     }
     res[i] = x[i] ^ y[i] ;
   }
 }
 
 for(int i = 0 ; i < n ; i++) 
   cout << y[i] << " " ;
}

int main() {
  fast_io
  cin >> t ;
  while(t--) {
    cin >> n ;
    x.resize(n) ;
    for(int i = 0 ; i < n ; i++) {
      cin >> x[i] ;
    }
    solve() ;
    cout << "\n" ;
  }
return 0 ;
}

```