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
1
5
6 5 12 5 16
 1
5
3 5 6 11 2

2
 0

import java.util.*;

public class C_Paint_the_Array{
    
    static int N = 110;
    static int n;
    static long[] a = new long[N];

    public static void main(String[] args){
    
        Scanner sc = new Scanner(System.in);
        int T = sc.nextInt();
        while(T-- > 0){
            n = sc.nextInt();
            for(int i = 1; i <= n; i++) a[i] = sc.nextLong();
            slover();
        }
    }

    static void slover(){
        long res = 0;

        // 情况1 frist one is divisible
        boolean flag1 = false;
        long cur = a[1];
        // System.out.print(cur + " ");
        for(int i = 3; i <= n; i += 2){
            cur = gcd(cur, a[i]);
            // System.out.print(a[i] + " ");
        }
        // System.out.println();
        // System.out.println("1st -> " + cur);

        // check other is divible or not
        for(int i = 2; i <= n; i += 2){
            if(a[i] % cur == 0){
                flag1 = true;
                break;
            }
        }
        if(!flag1) res = cur;

        // 情况2 second one is divisible
        if(flag1){
            boolean flag2 = false;
            cur = a[2];
            // System.out.print(cur + " ");
            for(int i = 4; i <= n; i+= 2){
                cur = gcd(cur, a[i]);
                // System.out.print(a[i] + " ");
            }
            // System.out.println();
            // System.out.println("2nd -> " + cur);
            for(int i = 1; i <= n; i+= 2){
                if(a[i] % cur == 0){
                    flag2 = true;
                    break;
                }
            }
            if(!flag2) res = cur;
        }

        System.out.println(res);
        // System.out.println("------");

    }

    static long gcd(long a, long b){
        if(b != 0) return gcd(b, a % b);
        else return a;
    }

}

/**
 * 10
 * 9 8 2 6 6 2 8 6 5 4
 * 
 * D题
 * n = 4,  k = 2
 * 1 3 3 7
 * 1. 先找最小的，最小的作为分子
 * 1. 大的一定要除了
 * 2. 剩余的也要加进去
 * 3. 
 * 3 
 * 1 1 1 2 1 3 1
 * 1 1 1 1 1 2 3 sorted
 * 
 * 9 2
 * 1 10 10 1 10 2 7 10 3
 * 1 1 2 3 7 10 10 10 10
 * 
 * 0
*/