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.util.*;
import java.io.*;
import java.lang.*;
public class codeforces {
    static PrintWriter pw = new PrintWriter(System.out);
    public static void main(String []args){
          Scanner sc = new Scanner(System.in);
          int t = sc.nextInt();
          while(t-->0) {
            int n = sc.nextInt();
            int a[] = new int[n];
              int b[] = new int[n];
            for(int i=0;i<n;i++)
                a[i] = sc.nextInt();

           for(int i=1;i<n;i++){
               int index=31;
               for(int j=31;j>=0;j--){
                   if((a[i-1]&((long)1<<j))==(1<<j)) {
                       index = j;
                       break;
                   }
               }
               int sum=0;
               for(int j=index;j>=0;j--){
                   if((a[i-1]&((long)1<<j))==(1<<j)&&(a[i]&((long)1<<j))==0){
                       sum+=(1<<j);
                       a[i]+=(1<<j);
                   }
                  b[i] = sum;
               }
           }
                for(int i : b)
                    System.out.print(i+" ");
                System.out.println();
          }
     }
     public static void union(int a,int b,int Parent[]){
        Parent[a] += Parent[b];
        Parent[b] = a;
     }
      public static int find(int a,int Parent[]){
         if(Parent[a]<0)
              return a;
         return (Parent[a] =find(Parent[a],Parent));
      }
    public static class Pair{
        public  int x;
        public  int y;
        public Pair(int x,int y){
            this.x=x;
            this.y=y;
        }
    }
    public static long gcd(long a,long b){

        if(a==0)
            return b;
        return gcd(b%a,a);
    }
    public static class FastScanner {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st=new StringTokenizer("");
        String next() {
            while (!st.hasMoreTokens())
                try {
                    st=new StringTokenizer(br.readLine());
                } catch (IOException e) {
                    e.printStackTrace();
                }
            return st.nextToken();
        }

        int nextInt() {
            return Integer.parseInt(next());
        }
        int[] readArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }
}
