#include <bits/stdc++.h>
using namespace std;
#define int long long
#define loop(i,x,y) for(int i=x; i<y; ++i)
#define pb push_back
#define mp make_pair
#define FAST_IO ios_base::sync_with_stdio(false); cin.tie(NULL);
typedef vector<int> vi;
typedef vector<vector<int> > vvi;
typedef pair<int,int> pii;
typedef vector<pair<int,int> > vpii;

struct vectorHasher {
	int operator()(const vi& v) const {
		int hash = v.size();
		for(auto& i:v) {
			hash ^= i + 0x9e3779b9 + (hash<<6) + (hash>>2);
		}
		return hash;
	}	
};


struct pairHasher {
	int operator()(const pii& p) const {
		return hash<int>()(p.first)^hash<int>()(p.second);
	}
};

signed main() {
	FAST_IO
	int t;
	cin >>t;
	while(t--) {
	    int n;
	    cin >> n;
	    vector<int> x(n);
	    for(auto& i:x) cin >> i;
	    bool flag = true;
	    for(int i=0; i<n-1; ++i) {
	        flag = flag && ((x[i] & x[i+1]) == x[i]);
	    }
	    vector<int> y = {0};
	    for(int i=0; i<n-1; ++i) {
	        if(x[i+1]==x[i]) {
	            y.push_back(0);
	            continue;
	        }
	        
	        int temp = 0;
	        int sz = (int)log2(x[i])+1;
	        for(int j=0; j<sz; ++j) {
	            if((x[i+1] & 1<<j) < (x[i] & 1<<j)) {
	                temp = temp | (1<<j);
	            }
	        }
	        y.push_back(temp);
	        x[i+1] = x[i+1] ^ y[i+1];
	    }
	    
	    for(auto i:y) cout << i << " ";
	    cout << endl;
	}
	
	
	return 0;
}

### Prompt

Your challenge is to write 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
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define loop(i,x,y) for(int i=x; i<y; ++i)
#define pb push_back
#define mp make_pair
#define FAST_IO ios_base::sync_with_stdio(false); cin.tie(NULL);
typedef vector<int> vi;
typedef vector<vector<int> > vvi;
typedef pair<int,int> pii;
typedef vector<pair<int,int> > vpii;

struct vectorHasher {
	int operator()(const vi& v) const {
		int hash = v.size();
		for(auto& i:v) {
			hash ^= i + 0x9e3779b9 + (hash<<6) + (hash>>2);
		}
		return hash;
	}	
};


struct pairHasher {
	int operator()(const pii& p) const {
		return hash<int>()(p.first)^hash<int>()(p.second);
	}
};

signed main() {
	FAST_IO
	int t;
	cin >>t;
	while(t--) {
	    int n;
	    cin >> n;
	    vector<int> x(n);
	    for(auto& i:x) cin >> i;
	    bool flag = true;
	    for(int i=0; i<n-1; ++i) {
	        flag = flag && ((x[i] & x[i+1]) == x[i]);
	    }
	    vector<int> y = {0};
	    for(int i=0; i<n-1; ++i) {
	        if(x[i+1]==x[i]) {
	            y.push_back(0);
	            continue;
	        }
	        
	        int temp = 0;
	        int sz = (int)log2(x[i])+1;
	        for(int j=0; j<sz; ++j) {
	            if((x[i+1] & 1<<j) < (x[i] & 1<<j)) {
	                temp = temp | (1<<j);
	            }
	        }
	        y.push_back(temp);
	        x[i+1] = x[i+1] ^ y[i+1];
	    }
	    
	    for(auto i:y) cout << i << " ";
	    cout << endl;
	}
	
	
	return 0;
}

```