human
Mocha is a young girl from high school. She has learned so much interesting knowledge from her teachers, especially her math teacher. Recently, Mocha is learning about binary system and very interested in bitwise operation.

This day, Mocha got a sequence a of length n. In each operation, she can select an arbitrary interval [l, r] and for all values i (0≤ i ≤ r-l), replace a_{l+i} with a_{l+i}  \&  a_{r-i} at the same time, where \& denotes the [bitwise AND operation](https://en.wikipedia.org/wiki/Bitwise_operation#AND). This operation can be performed any number of times.

For example, if n=5, the array is [a_1,a_2,a_3,a_4,a_5], and Mocha selects the interval [2,5], then the new array is [a_1,a_2 \&  a_5, a_3 \&  a_4, a_4 \&  a_3, a_5 \&  a_2].

Now Mocha wants to minimize the maximum value in the sequence. As her best friend, can you help her to get the answer?

Input

Each test contains multiple test cases. 

The first line contains a single integer t (1 ≤ t ≤ 100) — the number of test cases. Each test case consists of two lines.

The first line of each test case contains a single integer n (1 ≤ n ≤ 100) — the length of the sequence.

The second line of each test case contains n integers a_1, a_2, …, a_n (0 ≤ a_i ≤ 10^9).

Output

For each test case, print one integer — the minimal value of the maximum value in the sequence.

Example

Input

4
2
1 2
3
1 1 3
4
3 11 3 7
5
11 7 15 3 7

Output

0
1
3
3

Note

In the first test case, Mocha can choose the interval [1,2], then the sequence becomes [ 0, 0], where the first element is 1 \& 2, and the second element is 2 \& 1.

In the second test case, Mocha can choose the interval [1,3], then the sequence becomes [ 1,1,1], where the first element is 1 \& 3, the second element is 1 \& 1, and the third element is 3 \& 1.
Using python3 can you solve the prior task?

Write me a Python program to print one integer — the minimal value of the maximum value in the sequence?
assistant
#Here is Python program to print one integer — the minimal value of the maximum value in the sequence.

import sys, os, io
def rs(): return sys.stdin.readline().rstrip()
def ri(): return int(sys.stdin.readline())
def ria(): return list(map(int, sys.stdin.readline().split()))
def ws(s): sys.stdout.write(s + '\n')
def wi(n): sys.stdout.write(str(n) + '\n')
def wia(a): sys.stdout.write(' '.join([str(x) for x in a]) + '\n')
import math,datetime,functools,itertools,operator,bisect,fractions,statistics
from collections import deque,defaultdict,OrderedDict,Counter
from fractions import Fraction
from decimal import Decimal
from sys import stdout
from heapq import heappush, heappop, heapify ,_heapify_max,_heappop_max,nsmallest,nlargest
# sys.setrecursionlimit(111111) 
INF=999999999999999999999999
alphabets="abcdefghijklmnopqrstuvwxyz"
class SortedList:
    def __init__(self, iterable=[], _load=200):
        """Initialize sorted list instance."""
        values = sorted(iterable)
        self._len = _len = len(values)
        self._load = _load
        self._lists = _lists = [values[i:i + _load] for i in range(0, _len, _load)]
        self._list_lens = [len(_list) for _list in _lists]
        self._mins = [_list[0] for _list in _lists]
        self._fen_tree = []
        self._rebuild = True
 
    def _fen_build(self):
        """Build a fenwick tree instance."""
        self._fen_tree[:] = self._list_lens
        _fen_tree = self._fen_tree
        for i in range(len(_fen_tree)):
            if i | i + 1 < len(_fen_tree):
                _fen_tree[i | i + 1] += _fen_tree[i]
        self._rebuild = False
 
    def _fen_update(self, index, value):
        """Update `fen_tree[index] += value`."""
        if not self._rebuild:
            _fen_tree = self._fen_tree
            while index < len(_fen_tree):
                _fen_tree[index] += value
                index |= index + 1
 
    def _fen_query(self, end):
        """Return `sum(_fen_tree[:end])`."""
        if self._rebuild:
            self._fen_build()
 
        _fen_tree = self._fen_tree
        x = 0
        while end:
            x += _fen_tree[end - 1]
       ...