C

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

331 1 152 2 2 3 241 2 4 8

Sample Output

12
2
这道题刚开始我就没有思路，看了一下别人的代码，原来还要用到前缀和后缀。思路如下：
依次遍历每个数，算出这堆数的左边最大公约数，然后求出这堆数右边的最大公约数，然后求出他们的共同的最大公约数，遍历每个最大公约数即为所求。
#include<iostream>#include<cstdio>#include<algorithm>using namespace std;const int N=100005;int GCD(int x,int y){   if(x==0)        return y;    else        return GCD(y%x,x);  }int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n,a[N],ar[N],al[N],MAX=0;        scanf("%d",&n);        for(int i=0;i<n;i++)        {          scanf("%d",&a[i]);          if(i==1)             al[i]=a[i];          else            al[i]=GCD(al[i-1],a[i]);        }        for(int i=n;i>0;i--)        {            if(i==n)                ar[i]=a[i];            else                ar[i]=GCD(ar[i+1],a[i]);        }        for(int i=1;i<=n;i++)        {            if(i==1)                MAX=max(MAX,ar[2]);            if(i==n)                MAX=max(MAX,al[n-1]);            else                MAX=max(MAX,GCD(al[i-1],ar[i+1]));        }        printf("%d\n",MAX);    }    return 0;}