Problem:
Score : 300 points

Problem Statement

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.

Sample Input 1

3
7 6 8

Sample Output 1

2

If we replace 7 with 4, the greatest common divisor of the three integers on the blackboard will be 2, which is the maximum possible value.

Sample Input 2

3
12 15 18

Sample Output 2

6

Sample Input 3

2
1000000000 1000000000

Sample Output 3

1000000000

We can replace an integer with itself.

Solution in Rust (1.15.1) (judge verdict: Accepted)

#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
#[allow(unused_imports)]
use std::io::{Write, BufWriter};
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, [ $t:tt ]) => {{
        let len = read_value!($next, usize);
        (0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    }};

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}
pub fn gcd(x: i64, y: i64) -> i64 {
    if y == 0 {
        x.abs()
    } else {
        gcd(y, x % y)
    }
}

fn solve() {
    let mo:i64 = 1_000_000_007;
    input! {
        n: usize,
        a: [i64; n],
    }
    let mut left = vec![0; n];
    let mut right = vec![0; n];
    left[0] = a[0];
    right[n-1] = a[n-1];
    for i in 1..n{
        left[i] = gcd(left[i-1], a[i]);
        right[n-i-1] = gcd(right[n-i], a[n-i-1]);
    }
    let mut ans = 0;
    ans = max(left[n-2],right[1]);
    for i in 0..n-2{
        ans = max(ans, gcd(left[i],right[i+2]));
    }
    println!("{}", ans);
}

fn main() {
    // In order to avoid potential stack overflow, spawn a new thread.
    let stack_size = 104_857_600; // 100 MB
    let thd = std::thread::Builder::new().stack_size(stack_size);
    thd.spawn(|| solve()).unwrap().join().unwrap();
}