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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <iomanip>

using namespace std;

#define PB push_back
#define ALL(v) (v).begin(), (v).end()
#define SZ size()
#define FOR(i, a, b) for(int (i) = (a); (i) < (b); (i)++)
#define REP(i, n) FOR((i), 0, (n))
#define GET(a) cin >> (a)
#define GET2(a, b) cin >> (a) >> (b)
#define GET3(a, b, c) cin >> (a) >> (b) >> (c)
#define SHOW(v) cout << (v) << endl
#define SHOW2(a, b) cout << (a) << " " << (b) << endl
#define SHOW3(a, b, c) cout << (a) << " " << (b) << " " << (c) << endl
#define MOD 1000000007
template<class T> void GETV(T &v) { REP(i, v.size()) { GET(v[i]); } }
template<class T> void SHOWV(T &v) { REP(i, v.size()) { SHOW(v[i]); } }

typedef long long ll;

int gcd(int a, int b) {
	if(a % b == 0) {
		return b;
	}
	else {
		return gcd(b, a % b);
	}
}

int main() {
	int n;
	cin >> n;
	
	vector<int> v(n);
	
	for(int i = 0; i < n; i++) {
		cin >> v[i];
	}
	
	int max = gcd(v[0], v[1]);
	for(int i = 2; i < n; i++) {
		max = gcd(max, v[i]);
	}
	
	int index1 = 0;
	int index2 = 0;
	int tmp = v[0];
	for(int i = 0; i < n; i++) {
		tmp = gcd(tmp, v[i]);
		if(tmp == max) {
			index1 = i - 1;
			index2 = i;
			break;
		}
	}
	
	int ans1 = 0;
	int ans2 = 0;
	for(int i = 0; i < n; i++) {
		if(i != index1) {
			if(ans1 == 0) {
				ans1 = v[i];
			}
			else {
				ans1 = gcd(ans1, v[i]);
			}
		}
		if(i != index2) {
			if(ans2 == 0) {
				ans2 = v[i];
			}
			else {
				ans2 = gcd(ans2, v[i]);
			}
		}
	}
	
	cout << (ans1 > ans2 ? ans1 : ans2) << endl;
	
    return 0;
}