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 
import java.io.*;
import java.util.*; 
public class AA {
    public static void main(String[] args) throws Exception {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = null;
        int t = 1;
        
        t = Integer.parseInt(br.readLine());

        while(t-->0) {
            int n = Integer.parseInt(br.readLine());
            st = new StringTokenizer(br.readLine());
            int[] x = new int[n];
            for(int i=0; i<n; i++) 
                x[i] = Integer.parseInt(st.nextToken());

            int[] y = new int[n];
            int cur = x[0];
            for(int i=1; i<n; i++) {
                cur = cur|x[i];
                y[i] = cur^x[i];
            }

            for(int i=0; i<n; i++) 
                System.out.print(y[i]+" ");
            System.out.println();
        }
    }


    static boolean[] isPrime;
    public static void sieve(int n) {
        isPrime = new boolean[n+1];
        Arrays.fill(isPrime, true);
        isPrime[0] = isPrime[1] = false;
        for(int i=2; i*i<=n; i++) {
            if(isPrime[i]) {
                for(int j=i*i; j<=n; j+=i) {
                    isPrime[j] = false;
                }
            }
        }
    }


    public static long pow(long a, long b ) {
        long res = 1;
        while(b>0) {
            if((b&1)==1)
                res *= a;
            a *= a;
            b >>= 1;
        }
        return res;
    }

    public static long gcd(long a, long b) {
        while (b!=0) {
            a %= b;
            long t = a;
            a = b;
            b = t;
        }
        return a;
    }

    public static long lcm(long a, long b) {
        return a*b/gcd(a,b);
    }
}

class Pair {
    int f;
    int s;
    Pair() {}
    Pair(int f, int s) {
        this.f = f;
        this.s = s;
    }
}