C

T. Chur teaches various groups of students at university U. Every U-student has a unique Student Identification Number (SIN). A SIN s is an integer in the range 0 ≤ s ≤ MaxSIN with MaxSIN = 10 ⁶-1. T. Chur finds this range of SINs too large for identification within her groups. For each group, she wants to find the smallest positive integer m, such that within the group all SINs reduced modulo m are unique.

Input

On the first line of the input is a single positive integer N, telling the number of test cases (groups) to follow. Each case starts with one line containing the integer G (1 ≤ G ≤ 300): the number of students in the group. The following G lines each contain one SIN. The SINs within a group are distinct, though not necessarily sorted.

Output

For each test case, output one line containing the smallest modulus m, such that all SINs reduced modulo m are distinct.

Sample Input

2
1
124866
3
124866
111111
987651

Sample Output

1
8


该题，整体思路有，不过会觉得肯定会超时，结果也超时了，后来搜了题解，有一个小小的技巧
就是memset的时候，不整体memset，而是从0到i+1，因为下一层循环对i取得膜，所以vis里面改变的下标都<=i！！！！！十分重要，不这样就会超时！！！

#include <iostream>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
using namespace std;
const int maxn=1e6+10;
bool vis[maxn];
int a[500];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        int g;
        scanf("%d",&g);
        for(int i=0;i<g;i++) scanf("%d",&a[i]);

        if(g==1) printf("1
");
        else
        {
            int i;
            for(i=2;;i++)
            {
                memset(vis,0,sizeof(bool)*(i+1));//！！！！
                bool flag=0;
                for(int j=0;j<g;j++)
                {
                    if(vis[a[j]%i]){flag=1;break;}
                    else vis[a[j]%i]=1;
                }
                if(flag==0) break;
            }
            printf("%d
",i);
        }
    }
}