G - coprime sequence HDU - 6025

Do you know what is called ``Coprime Sequence''? That is a sequence consists of nnpositive integers, and the GCD (Greatest Common Divisor) of them is equal to 1. 
``Coprime Sequence'' is easy to find because of its restriction. But we can try to maximize the GCD of these integers by removing exactly one integer. Now given a sequence, please maximize the GCD of its elements.

Input

The first line of the input contains an integer T(1≤T≤10)T(1≤T≤10), denoting the number of test cases. 
In each test case, there is an integer n(3≤n≤100000)n(3≤n≤100000) in the first line, denoting the number of integers in the sequence. 
Then the following line consists of nn integers a1,a2,...,an(1≤ai≤109)a1,a2,...,an(1≤ai≤109), denoting the elements in the sequence.

Output

For each test case, print a single line containing a single integer, denoting the maximum GCD.

Sample Input

3
3
1 1 1
5
2 2 2 3 2
4
1 2 4 8

Sample Output

1
2
2

#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;
int a[100000+10];
int b[100000+10];
int gcd(int x,int y){
	return y==0?x:(gcd(y,x%y));
}
int main()
{
	int t;
	cin>>t;
	while(t--){
		int n;
		cin>>n;
		memset(a,0,sizeof(a));
		memset(b,0,sizeof(b));
		for(int i=0;i<n;i++){
			cin>>a[i];
		}
		int k,l=0,j,s1,M=0;
		b[0]=b[1]=-1;
		for(int i=0;i<n;i++){
			k=a[0];
			if(i==0) k=a[1];
			for(j=0;j<n;j++){
				if(i==j) continue;
				k=gcd(a[j],k);
				if(k<=M)  break;//优化。。。
                if(j==i+1){
                    if(k==b[j])
                    break;
                }
                if(j==i+2)  b[j]=k;//if(j==n-1) b[l++]=k;//这样会超时
			}
			if(M<k)  M=k;
		}
		//sort(b,b+l);
		//cout<<b[l-1]<<endl;
		cout<<M<<endl;
	}
	return 0;
}