You will be given two sets of integers. Let‘s call them set A and set B. Set A contains n elements and set B contains m elements. You have to remove k₁ elements from set A and k₂ elements from set B so that of the remaining values no integer in set B is a multiple of any integer in set A. k₁ should be in the range [0, n] and k₂ in the range [0, m].

You have to find the value of (k₁ + k₂) such that (k₁ + k₂) is as low as possible. P is a multiple of Q if there is some integer K such that P = K * Q.

Suppose set A is {2, 3, 4, 5} and set B is {6, 7, 8, 9}. By removing 2 and 3 from A and 8 from B, we get the sets {4, 5} and {6, 7, 9}. Here none of the integers 6, 7 or 9 is a multiple of 4 or 5.

So for this case the answer is 3 (two from set A and one from set B).

Input

Input starts with an integer T (≤ 50), denoting the number of test cases.

The first line of each case starts with an integer n followed by n positive integers. The second line starts with m followed by m positive integers. Both n and mwill be in the range [1, 100]. Each element of the two sets will fit in a 32 bit signed integer.

Output

For each case of input, print the case number and the result.

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <string>
#include <algorithm>
#define LL long long
#define MAXN 100+10
#define MAXM 20000+10
#define INF 0x3f3f3f3f
using namespace std;
int a[110], b[110];
int Map[110][110];
bool used[110];
int pipei[110];
int N, M;
int k = 1;
int find(int x)
{
    for(int i = 1; i <= N; i++)
    {
        if(!used[i] && Map[x][i])
        {
            used[i] = true;
            if(pipei[i] == -1 || find(pipei[i]))
            {
                pipei[i] = x;
                return 1;
            }
        }
    }
    return 0;
}
void solve()//求匹配
{
    int ans = 0;
    memset(pipei, -1, sizeof(pipei));
    for(int i = 1; i <= M; i++)
    {
        memset(used, false, sizeof(used));
        ans += find(i);
    }
    printf("Case %d: ", k++);
    printf("%d\n", ans);
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        memset(Map, 0, sizeof(Map));
        scanf("%d", &N);
        for(int i = 1; i <= N; i++)
            scanf("%d", &a[i]);
        scanf("%d", &M);
        for(int i = 1; i <= M; i++)
        {
            scanf("%d", &b[i]);
            for(int j = 1; j <= N; j++)
            {
                if(b[i] % a[j] == 0)//整除关系建边
                    Map[i][j] = 1;
            }
        }
        solve();
    }
    return 0;
}