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 java.math.*;
  import java.util.*;
  import java.io.*;
  public class Template {
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader() {
              br = new BufferedReader(new
                      InputStreamReader(System.in));
          }
  
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          String nextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  
      public static long gcd(long a, long b) { return b==0 ? a : gcd(b, a%b); }
  
      static boolean checkEven(long even, long arr[]){
          for (int i = 1; i <arr.length ; i+=2) {
              if(arr[i]%even==0)
                  return false;
          }
          return true;
      }
      static boolean checkOdd(long odd, long arr[]){
          for (int i = 0; i <arr.length ; i+=2) {
              if(arr[i]%odd==0)
                  return false;
          }
          return true;
      }
  
  
      static void solve(int n, long arr[]){
          long odd=arr[1],even=arr[0];
          for (int i = 2; i < n; i++) {
              if(i%2==0)
                  even=gcd(even,arr[i]);
              else
                  odd=gcd(odd,arr[i]);
          }
          if(checkEven(even,arr))
              System.out.println(even);
          else if(checkOdd(odd,arr))
              System.out.println(odd);
          else System.out.println(0);
      }
  
      public static void main(String[] args) {
          FastReader sc = new FastReader();
          int t=sc.nextInt();
          for (int i = 0; i < t; i++) {
              int n=sc.nextInt();
              long arr[]=new long[n];
              for (int j = 0; j < n; j++) {
                  arr[j]=sc.nextLong();
              }
              solve(n,arr);
          }
      }
  
  
  }
  
  
  