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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;

/** Built using CHelper plug-in Actual solution is at the top */
public class Main {
  public static void main(String[] args) {
    InputStream inputStream = System.in;
    OutputStream outputStream = System.out;
    FastScanner in = new FastScanner(inputStream);
    PrintWriter out = new PrintWriter(outputStream);
    DCoGrowingSequence solver = new DCoGrowingSequence();
    int testCount = Integer.parseInt(in.next());
    for (int i = 1; i <= testCount; i++) solver.solve(i, in, out);
    out.close();
  }

  static class DCoGrowingSequence {
    public void solve(int testNumber, FastScanner in, PrintWriter out) {
      int n = in.nextInt();
      long[] a = in.nextLongArray(n);
      long[] ans = new long[n];
      for (int i = 1; i < n; ++i) {
        long cur = a[i];
        long pre = a[i - 1];
        long y = 0;
        for (long bit = 0; bit < 32; ++bit) {
          if (((1L << bit) & pre) != 0 && ((1L << bit) & cur) == 0) {
            y |= (1L << bit);
          }
        }
        a[i] ^= y;
        ans[i] = y;
      }
      ArrayUtils.printArray(out, ans);
    }
  }

  static class ArrayUtils {
    public static void printArray(PrintWriter out, long[] array) {
      if (array.length == 0) return;
      for (int i = 0; i < array.length; i++) {
        if (i != 0) out.print(" ");
        out.print(array[i]);
      }
      out.println();
    }
  }

  static class FastScanner {
    BufferedReader reader;
    StringTokenizer tokenizer;

    public FastScanner(InputStream inputStream) {
      reader = new BufferedReader(new InputStreamReader(inputStream), 32768);
      tokenizer = null;
    }

    public String next() {
      while (tokenizer == null || !tokenizer.hasMoreTokens()) {
        try {
          tokenizer = new StringTokenizer(reader.readLine());
        } catch (IOException e) {
          throw new RuntimeException(e);
        }
      }
      return tokenizer.nextToken();
    }

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

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

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