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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<math.h>
long gcd(long a,long b){
 // printf("%ld %ld\n",a,b);
if(a<b){
  long w=a;
  a=b;
  b=w;
}
  
  if(a%b!=0)return gcd(b,a%b);
  else return b;
}
int main(void){
int n;
  scanf("%d",&n);
  long a[n];
  for(int i=0;i<n;i++){
   	scanf("%ld",&a[i]);
  }
  long l[n],r[n];
  l[0]=a[0];
 // printf("%ld ",l[0]);
  for(int i=1;i<n;i++){
	l[i]=gcd(l[i-1],a[i]);
   // printf("%ld ",l[i]);
  }
  //printf("\n");
  r[n-1]=a[n-1];
  //printf("%ld ",r[n-1]);
  for(int i=1;i<n;i++){
	r[n-1-i]=gcd(r[n-i],a[n-1-i]);
  //  printf("%ld ",r[n-1-i]);
  }
 // printf("\n");
  long max=1000000001;
  for(int i=0;i<n;i++){
    long yaku;
    if(i==0)yaku=r[i+1];
    else if(i==n-1)yaku=l[n-1-i];
    else yaku=gcd(l[i-1],r[i+1]);
    if(yaku<max)max=yaku;
  }
  printf("%ld",max);
  
      return 0;
    }