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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;

public class D {
    private static FastReader fr = new FastReader();
    private static PrintWriter out=new PrintWriter(System.out);
    private static Random random = new Random();

    public static void main(String[] args) throws IOException {
        StringBuilder sb = new StringBuilder();
        // code goes here
        int t = fr.nextInt();
        while (t-- > 0){
            int n = fr.nextInt();
            int[] a = fr.nextIntArray(n);
            int[] b = new int[n];
            b[0] = 0;
            for(int i = 1; i < n; i++){
                int temp = b[i - 1] ^ a[i - 1];
                int t2 = a[i];
                int c = 0;
                int l = 0;
                while (temp > 0){
                    int bit = temp & 1;
                    int req = t2 & 1;
                    if(bit == 1 && req == 0){
                        c += (1 << l);
                    }
                    l++;
                    temp/=2;
                    t2/=2;
                }
                b[i] = c;
            }
            for(int i = 0; i < n; i++){
                sb.append(b[i]).append(" ");
            }
            sb.append("\n");
        }
        System.out.print(sb.toString());
    }

    static void ruffleSort(int[] a) {
        int n=a.length;//shuffle, then sort
        for (int i=0; i<n; i++) {
            int oi=random.nextInt(n), temp=a[oi];
            a[oi]=a[i]; a[i]=temp;
        }
        Arrays.sort(a);
    }

    static class FastReader{
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");

        public String next() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public int[] nextIntArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }

        public long nextLong() {
            return Long.parseLong(next());
        }

        public long[] nextLongArray(int n) {
            long[] a=new long[n];
            for (int i=0; i<n; i++) a[i]=nextLong();
            return a;
        }
    }

    static class Pair<A, B>{
        A first;
        B second;
        public Pair(A first, B second){
            this.first = first;
            this.second = second;
        }
    }

    static long mod(String num, long a)
    {
        // Initialize result
        long res = 0;

        // One by one process all digits of 'num'
        for (int i = 0; i < num.length(); i++)
            res = (res*10 +  num.charAt(i) - '0') %a;

        return res;
    }

    static long binomialCoeff(long n, long k, long MOD)
    {
        long res = 1;

        // Since C(n, k) = C(n, n-k)
        if (k > n - k)
            k = n - k;

        // Calculate value of
        // [n * (n-1) *---* (n-k+1)] / [k * (k-1) *----* 1]
        for (int i = 0; i < k; ++i) {
            res *= (n - i);
            res /= (i + 1);
            res %= MOD;
        }

        return res;
    }

    static long power(long x, long y, long p)
    {

        // Initialize result
        long res = 1;

        // Update x if it is more than or
        // equal to p
        x = x % p;

        while (y > 0) {

            // If y is odd, multiply x
            // with result
            if (y % 2 == 1)
                res = (res * x) % p;

            // y must be even now
            y = y >> 1; // y = y/2
            x = (x * x) % p;
        }

        return res;
    }

    // Returns n^(-1) mod p
    static long modInverse(long n, long p)
    {
        return power(n, p - 2, p);
    }

    // Returns nCr % p using Fermat's
    // little theorem.
    static long nCrModPFermat(int n, int r,
                              long p)
    {

        // Base case
        if (r == 0)
            return 1;

        // Fill factorial array so that we
        // can find all factorial of r, n
        // and n-r
        long[] fac = new long[n + 1];
        fac[0] = 1;

        for (int i = 1; i <= n; i++)
            fac[i] = fac[i - 1] * i % p;

        return (fac[n] * modInverse(fac[r], p)
                % p * modInverse(fac[n - r], p)
                % p)
                % p;
    }
}
