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 collections
import math
import sys
from functools import reduce
import decimal

sys.setrecursionlimit(10 ** 9)
decimal.getcontext().rounding = decimal.ROUND_HALF_UP

graphDict = collections.defaultdict
queue = collections.deque()


class Graphs:
    def __init__(self):
        self.graph = graphDict(list)

    def add_edge(self, u, v):
        self.graph[u].append(v)

    def dfs_utility(self, nodes, visited_nodes):
        visited_nodes.add(nodes)
        for neighbour in self.graph[nodes]:
            if neighbour not in visited_nodes:
                self.dfs_utility(neighbour, visited_nodes)
            else:
                return neighbour

    def dfs(self, node):
        Visited = set()
        self.dfs_utility(node, Visited)

    def bfs(self, node):
        visited = set()
        if node not in visited:
            queue.append(node)
            visited.add(node)
        while queue:
            parent = queue.popleft()
            print(parent)
            for item in self.graph[parent]:
                if item not in visited:
                    queue.append(item)
                    visited.add(item)


def rounding(n):
    return int(decimal.Decimal(f'{n}').to_integral_value())


def factors(n):
    return set(reduce(list.__add__,
                      ([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))


def inp():
    return sys.stdin.readline().strip()


def map_inp(v_type):
    return map(v_type, inp().split())


def list_inp(v_type):
    return list(map_inp(v_type))


######################################## Solution ####################################

for _ in range(int(inp())):
    x = int(inp())
    arr = list_inp(int)
    if x == 1:
        print(0)
    else:
        first = arr[0]
        answer = [0]
        for item in arr[1:]:
            if item & first == first:
                first = item
                answer.append(0)
            else:
                num = (item | first)  ^ item
                answer.append(num)
                first = num ^ item
        print(*answer)
