description: 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.
solution: """
        ___.   .__    .__       .__            __    __       _________ 
  _____ \_ |__ |  |__ |__| _____|  |__   ____ |  | _|  | __   \______  \
  \__  \ | __ \|  |  \|  |/  ___/  |  \_/ __ \|  |/ /  |/ /       /    /
   / __ \| \_\ \   Y  \  |\___ \|   Y  \  ___/|    <|    <       /    / 
  (____  /___  /___|  /__/____  >___|  /\___  >__|_ \__|_ \_____/____/  
       \/    \/     \/        \/     \/     \/     \/    \/_____/       
  """
  import sys, io, os
  
  BUFSIZE = 8192
  
  
  class FastIO(io.IOBase):
      newlines = 0
  
      def __init__(self, file):
          self._fd = file.fileno()
          self.buffer = io.BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
  
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
  
      def readline(self):
          while self.newlines == 0:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
  
      def flush(self):
          if self.writable:
              os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
  
  
  class IOWrapper(io.IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
  
  
  def print(*args, **kwargs):
      """Prints the values to a stream, or to sys.stdout by default."""
      sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
      at_start = True
      for x in args:
          if not at_start:
              file.write(sep)
          file.write(str(x))
          at_start = False
      file.write(kwargs.pop("end", "\n"))
      if kwargs.pop("flush", False):
          file.flush()
  
  
  sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  # endregion
  MOD = 1000000007
  
  
  # from collections import defaultdict as dd,Counter,deque
  def si(): return input()
  
  
  def ii(): return int(input())
  
  
  def li(): return list(map(int, input().split()))
  
  
  def mi(): return map(int, input().split())
  
  
  def sout(v): print(v, end=' ')
  
  
  def d2b(n): return bin(n).replace("0b", "")
  
  
  def twod(n, m, num): return [[num for x in range(m)] for y in range(n)]
  
  
  def vow(): return ['a', 'e', 'i', 'o', 'u']
  
  
  def let(): return [chr(i) for i in range(97, 123)]
  
  
  def gcd(x, y):
      while y:
          x, y = y, x % y
      return x
  
  
  def ispow2(x):
      return (x and (not (x & (x - 1))))
  
  
  def prime_factors(n):
      i = 2
      factors = []
      while i * i <= n:
          if n % i:
              i += 1
          else:
              n //= i
              factors.append(i)
      if n > 1: factors.append(n)
      return (list(factors))
  
  
  def binary_search(arr, low, high, x):
      if high >= low:
          mid = (high + low) // 2
          if arr[mid] == x:
              return mid
          elif arr[mid] > x:
              return binary_search(arr, low, mid - 1, x)
          else:
              return binary_search(arr, mid + 1, high, x)
      else:
          return -1
  
  
  def sieve(n):
      primes = []
      least_prime_divs = [0] * (n + 1)
      for i in range(2, n + 1):
          if not least_prime_divs[i]:
              least_prime_divs[i] = i
              primes.append(i)
          for p in primes:
              if i * p > n or p > least_prime_divs[i]: break
              least_prime_divs[i * p] = p
      return primes
  
  
  t = ii()
  while t:
      t -= 1
      n=ii()
      a=li()
      mx=-1
      d={}
      for i in a:
          for j in range(34):
              if i&(1<<j):
                  mx=max(mx,j)
              else:
                  d[j]=True
      ans=0
      for i in range(mx+1):
          if i not in d: ans+=pow(2,i)
      print(ans)