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.ArrayList;
import java.util.StringTokenizer;

public class CoGrowing {

    //Fast Input
    static class FastReader {
        BufferedReader br;
        StringTokenizer st;

        public FastReader() {
            br = new BufferedReader(new InputStreamReader(System.in));
        }

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

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

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

        double nextDouble() {
            return Double.parseDouble(next());
        }

        String nextLine() {
            String str = "";
            try {
                str = br.readLine();
            } catch (IOException e) {
                e.printStackTrace();
            }
            return str;
        }
    }

    //Printing an array
    static void print(long[][] a, int r, int c) {
        PrintWriter pw = new PrintWriter(System.out);
        for (int i = 0; i < r; i++) {
            for (int j = 0; j < c; j++) {
                pw.print(a[i][j] + " ");
            }
            pw.println();
        }
        pw.flush();
    }

    //Used to generate random test cases
    //Must be changed according to input format
    static void testCaseGenerator(int t, int lowN, int highN, long lowA, long highA) throws FileNotFoundException {
        File input = new File("C:\\Users\\debar\\OneDrive\\Documents\\input.txt");
        PrintWriter pw = new PrintWriter(input);
        pw.println(t);
        for (int i = 0; i < t; i++) {
            int n = (int) (Math.random() * (highN - lowN)) + lowN;
            pw.println(n);
            for (int j = 0; j < n; j++) {
                long a = (long) (Math.random() * (highA - lowA)) + lowA;
                pw.print(a + " ");
            }
            pw.println();
        }
        pw.flush();
    }

    //Find factorial of numbers from 1 to n
    //Check the size of factorial array
    static long[] factorial(int n, long mod) {
        long[] factorial = new long[n];
        factorial[0] = 1;
        for (int i = 1; i <= n; i++) {
            factorial[i] = (factorial[i - 1] * i) % mod;
        }
        return factorial;
    }

    //Find factorial inverse from 1 to n
    //Check the size of inverse_factorial array
    static long[] factorialInverse(int n, long mod) {
        long[] inverse_factorial = modularInverse(n, mod);
        for (int i = 2; i <= n; i++) {
            inverse_factorial[i] = (inverse_factorial[i] * inverse_factorial[i - 1]) % mod;
        }
        return inverse_factorial;
    }

    //Inverse exist when two number are relatively prime
    //Based on Fermat little theorem (a^mod-1)%mod=1 when mod is prime
    //a * a^(mod - 2) % mod = 1
    //a^(mod - 2) % mod =a^-1
    //When mod != prime use Extended euclidean theorem to calculate x and y for equation a*x + mod*y = gcd(a,b)=1
    //(a*x + mod*y = gcd(a,b)=1) % mod = a*x %mod =1
    //https://www.geeksforgeeks.org/find-initial-integral-solution-of-linear-diophantine-equation-if-finite-solution-exists/?ref=rp
    //Used for only one number
    static long modularInverse(long a, long mod) {
        return binaryExponentiation(a, mod - 2, mod);
    }

    //Modular inverse for number from 1 to n - 1
    //Time complexity O(n) for n numbers
    //Proof https://cp-algorithms.com/algebra/module-inverse.html
    static long[] modularInverse(int n, long mod) {
        long[] inv = new long[n];
        inv[1] = 1;
        for (int i = 2; i < n; i++) {
            inv[i] = mod - (mod / i) * inv[(int) (mod % i)] % mod;
        }
        return inv;
    }

    //GCD calculation of a and b with complexity O(log(Max(a,b))
    static long gcd(long a, long b) {
        long r;
        while (b != 0) {
            r = a % b;
            a = b;
            b = r;
        }
        return a;
    }

    //Find prime number upto n with complexity O(n*log(log(n)))
    //Travel through the multiple of prime number and set it to false
    static boolean[] sieveOfEratosthenes(int n) {

        boolean[] prime = new boolean[n + 1];
        for (int i = 0; i <= n; i++)
            prime[i] = true;

        for (int p = 2; p * p <= n; p++) {
            if (prime[p]) {
                for (int i = p * p; i <= n; i += p)
                    prime[i] = false;
            }
        }

        prime[0] = prime[1] = false;
        return prime;
    }

    //Find Prefix sum in O(n)
    static long[] PrefixSum(long[] arr, int n) {

        long[] prefixSum = new long[n];

        prefixSum[0] = arr[0];
        for (int i = 1; i < n; ++i)
            prefixSum[i] = prefixSum[i - 1] + arr[i];
        return prefixSum;
    }

    //Find a^n in O(log n)
    //If n=11 then n in binary is equal to 1011
    //a^n=a^(8+2+1) that is multiply with result when bit is set.
    //https://youtu.be/L-Wzglnm4dM
    static long binaryExponentiation(long a, long n, long mod) {
        long result = 1L;
        n = n % (mod - 1);
        //n = (mod - 1) * floor(n / (mod - 1)) + n % (mod - 1)
        //a^((mod - 1) * floor(n / (mod - 1))) % mod = 1 so a^n % mod =a^(n % (mod - 1)) % mod when mod is prime
        //Using fermat little theorem a^(p-1) % p = 1 when p is prime
        while (n > 0) {
            if ((n & 1) == 1) {
                result = (result * a) % mod;
            }
            a = (a * a) % mod;
            n /= 2;
        }
        return result;
    }

    //Matrix multiplication
    static long[][] matrixMultiply(long[][] a, long[][] b, int row, int col, int mid, long mod) {
        long[][] c = new long[row][col];
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                c[i][j] = 0;
                for (int k = 0; k < mid; k++) {
                    c[i][j] = (c[i][j] + (a[i][k] * b[k][j]) % mod) % mod;
                }
            }
        }
        return c;
    }

    //Matrix Exponentiation
    static long[][] matrixExponentiation(long[][] a, long n, int size, long mod) {
        if (n == 0) {
            long[][] temp = new long[size][size];
            for (int i = 0; i < size; i++) {
                temp[i][i] = 1;
            }
            return temp;//Returns identity matrix
        } else {
            long[][] temp = matrixExponentiation(a, n / 2, size, mod);
            if (n % 2 == 0) {
                return matrixMultiply(temp, temp, size, size, size, mod);
            } else {
                return matrixMultiply(matrixMultiply(temp, temp, size, size, size, mod), a, size, size, size, mod);
            }
        }
    }

    //Solution starts here
    public static void main(String[] args) {

        //Input output class reference
        FastReader sc = new FastReader();
        PrintWriter pw = new PrintWriter(System.out);

        //Number of test case
        int t = sc.nextInt();

        //For each test case
        for (int i = 0; i < t; i++) {
            ArrayList<Long> y=new ArrayList<>();

            int n=sc.nextInt();
            long x[]=new long[n];

            for(int j=0;j<n;j++)
            {
                x[j]=sc.nextLong();
            }

            y.add(0L);

            for(int j=1;j<n;j++)
            {
                if(((x[j-1]^y.get(j-1)) & x[j]) !=(x[j-1]^y.get(j-1)))
                    y.add((x[j]|(x[j-1]^y.get(j-1)))^x[j]);
                else
                    y.add(0L);
            }

            for(long j:y)
            {
                pw.print(j+" ");
            }
            pw.println();
        }
        pw.flush();
    }
    //Solution ends here
}