There are N integers, A_1, A_2, ..., A_N, written on the blackboard.

You will choose one of them and replace it with an integer of your choice between 1 and 10^9 (inclusive), possibly the same as the integer originally written.

Find the maximum possible greatest common divisor of the N integers on the blackboard after your move.

Constraints

* All values in input are integers.
* 2 \leq N \leq 10^5
* 1 \leq A_i \leq 10^9

Output

Input is given from Standard Input in the following format:


N
A_1 A_2 ... A_N

Output

Print the maximum possible greatest common divisor of the N integers on the blackboard after your move.

Examples

Input

3
7 6 8


Output

2


Input

3
12 15 18


Output

6


Input

2
1000000000 1000000000


Output

1000000000
/*
 *ver3
 *l_gcdとr_gcdの配列計算方法変更
 */

import java.util.Scanner;
import java.util.Arrays;

public class Main{
    public static void main(String[] args){
	Scanner scan = new Scanner(System.in);

	int n = scan.nextInt();
	int[] a = new int[n];
	for(int i = 0; i < n; i++){
	    a[i] = scan.nextInt();
	}
	
	//最大公約数の配列
	int[] gcd = new int[n];

	//端の処理
	gcd[0] = ruisekiGcd(a, 1, n);
	gcd[n-1] = ruisekiGcd(a, 0, n-1);

	//右側、左側累積GCDを作成
	int[] l_gcd = new int[n];
	int[] r_gcd = new int[n];
	l_gcd[0] = a[0];
	r_gcd[n-1] = a[n-1];
	for(int i = 1; i < n-1; i++){
	    l_gcd[i] = getGcd(l_gcd[i-1], a[i-1]);
	    r_gcd[n-i-1] = getGcd(r_gcd[n-i], a[n-i]);
	}
	
	//累積GCDを作成
	for(int i = 1; i < n-1; i++){
	    gcd[i] = getGcd(l_gcd[i], r_gcd[i]);
	}

	//累積GCDの最大値を計算
	Arrays.sort(gcd);
	
	System.out.println(gcd[n-1]);
    }

    static int ruisekiGcd(int[] a, int s, int l){
	int gcd = a[s];
	for(int i = s+1; i < l; i++){
	    gcd = getGcd(gcd, a[i]);
	}
	return gcd;
    }

    static int getGcd(int a, int b){
	if(a < b){
	    int tmp = a;
	    a = b;
	    b = tmp;
	}

	while(b != 0){
	    int tmp = a % b;
	    a = b;
	    b = tmp;
	}
	
	return a;
    }
}
