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 
from itertools import combinations_with_replacement 
import sys
from sys import stdin
import math
import bisect
#Find Set LSB = (x&(-x)), isPowerOfTwo = (x & (x-1))
# 1<<x =2^x
#x^=1<<pos  flip the bit at pos

def BinarySearch(a, x):
    i = bisect.bisect_left(a, x)
    if i != len(a) and a[i] == x:
        return i
    else:
        return -1
def iinput():
    return int(input())
def minput():
    return map(int,input().split())
def linput():
    return list(map(int,input().split()))

def fiinput():
    return int(stdin.readline())
def fminput():
    return map(int,stdin.readline().strip().split())
def flinput():
    return list(map(int,stdin.readline().strip().split()))


for _ in range(iinput()):
    n=iinput()
    list1=linput()
    list2=[0]
    for i in range(1,n):
        prev=bin(list1[i-1])
        
        nex=bin(list1[i])
        l1=len(prev)
        l2=len(nex)
        if(l2<l1):
            nex=nex[0:2]+"0"*(l1-l2)+nex[2:]
        if(l1<l2):
            prev=prev[0:2]+"0"*(l2-l1)+prev[2:]
        # print(prev,nex)
        y="0b"
        for j in range(2,len(prev)):
            if(prev[j]=="0"):
                y=y+"0"
            elif(prev[j]=="1" and nex[j]=="1"):
                y=y+"0"
            elif(prev[j]=="1" and nex[j]=="0"):
                y=y+"1"
        list1[i]=list1[i]^int(y,2)
        # print(y)
        list2.append(int(y,2))
    print(*list2)
        
        
    #     print(x)
    #     y="0b"
    #     for j in range(2,len(x)):
    #         if(x[j]=="0"):
    #             y=y+"1"
    #         else:
    #             y=y+"0"
    #     y=int(y,2)
    #     list2.append(y)
    # print(*list2)
    
        
        
        
        
        
        
        
        
        
        