description: You are given an array a consisting of n positive integers. You have to choose a positive integer d and paint all elements into two colors. All elements which are divisible by d will be painted red, and all other elements will be painted blue.
  
  The coloring is called beautiful if there are no pairs of adjacent elements with the same color in the array. Your task is to find any value of d which yields a beautiful coloring, or report that it is impossible.
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of testcases.
  
  The first line of each testcase contains one integer n (2 ≤ n ≤ 100) — the number of elements of the array.
  
  The second line of each testcase contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^{18}).
  
  Output
  
  For each testcase print a single integer. If there is no such value of d that yields a beautiful coloring, print 0. Otherwise, print any suitable value of d (1 ≤ d ≤ 10^{18}).
  
  Example
  
  Input
  
  
  5
  5
  1 2 3 4 5
  3
  10 5 15
  3
  100 10 200
  10
  9 8 2 6 6 2 8 6 5 4
  2
  1 3
  
  
  Output
  
  
  2
  0
  100
  0
  3
solution: import os
  import sys
  from bisect import bisect, bisect_left, insort, insort_left
  from heapq import heapify, heappop, heappush, nlargest, nsmallest
  from io import BytesIO, IOBase
  from math import inf, sqrt
  from types import GeneratorType
  
  # Fast IO for Python
  
  BUFSIZE: int = 8192
  
  
  class FastIO(IOBase):
      newlines = 0
  
      def __init__(self, file):
          self._fd = file.fileno()
          self.buffer = 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(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)
  def input(): return sys.stdin.readline().rstrip("\r\n")
  
  
  # Bootstrap used for recursion, use yield instead of return
  
  def bootstrap(f, stack=[]):
      def wrappedfunc(*args, **kwargs):
          if stack:
              return f(*args, **kwargs)
          else:
              to = f(*args, **kwargs)
              while True:
                  if type(to) is GeneratorType:
                      stack.append(to)
                      to = next(to)
                  else:
                      stack.pop()
                      if not stack:
                          break
                      to = stack[-1].send(to)
              return to
      return wrappedfunc
  
  
  # Never use += for string concatenation
  # Always use .join for string concatenation
  # Use PyPy3 for Codeforces
  # Use Python3 for Codechef
  
  # Function to compute inverse of a under modulo m
  
  def modInverse(a, m):
  
      g = gcd(a, m)
  
      if (g != 1):
          return -1
  
      else:
          # If a and m are relatively prime,
          # then modulo inverse is a^(m-2) mode m
          return power(a, m - 2, m)
  
  
  # Function to compute x^y under modulo m
  
  def power(x, y, m):
  
      if (y == 0):
          return 1
  
      p = power(x, y // 2, m) % m
      p = (p * p) % m
  
      if(y % 2 == 0):
          return p
      else:
          return ((x * p) % m)
  
  
  # Function to return gcd of a and b
  
  def gcd(a, b):
      if (a == 0):
          return b
  
      return gcd(b % a, a)
  
  
  # Sieve of Eratosthenes to get prime upto n
  
  def SieveOfEratosthenes(n):
      prime = [True for i in range(n+1)]
      p = 2
      while (p * p <= n):
  
          # If prime[p] is not
          # changed, then it is a prime
          if (prime[p] == True):
  
              # Update all multiples of p
              for i in range(p * p, n+1, p):
                  prime[i] = False
          p += 1
  
      ans = set()
      for p in range(2, n+1):
          if prime[p]:
              ans.add(p)
  
      return ans
  
  
  MOD = 1000000007
  INT_MAX = inf
  INT_MIN = -inf
  
  
  # Main function
  
  def main():
      t = int(input())
      for _ in range(t):
          n = int(input())
          a = list(map(int, input().split()))
          g1, g2 = a[0], a[1]
          for i in range(0, n, 2):
              g1 = gcd(g1, a[i])
          for i in range(1, n, 2):
              g2 = gcd(g2, a[i])
  
          if (g1 == 1 and g2 == 1) or g1 == g2:
              print(0)
          else:
              # if gcd(g1, g2) == 1:
              #     print(max(g1, g2))
              # else:
              #     print(0)
  
              if g2 > g1:
                  check = False
                  for i in range(0, n, 2):
                      if a[i] % g2 == 0:
                          check = True
                          break
                  if check == True:
                      if g2 % g1 == 0:
                          print(0)
                      else:
                          check = False
                          for i in range(1, n, 2):
                              if a[i] % g1 == 0:
                                  check = True
                                  break
                          if check == True:
                              print(0)
                          else:
                              print(g1)
  
                  else:
                      print(g2)
  
              else:
                  check = False
                  for i in range(1, n, 2):
                      if a[i] % g1 == 0:
                          check = True
                          break
                  if check == True:
                      if g1 % g2 == 0:
                          print(0)
                      else:
                          check = False
                          for i in range(0, n, 2):
                              if a[i] % g2 == 0:
                                  check = True
                                  break
                          if check == True:
                              print(0)
                          else:
                              print(g2)
                  else:
                      print(g1)
  
  
  main()
  