/*

     模拟水题，排序后找出重复的ip就可以了

 */

 #include <cstdio>

 #include <iostream>

 #include <algorithm>

 #include <cstring>

 #include <string>

 #include <cmath>

 using namespace std;

 const int MAXN = 1e3 + ;

 const int INF = 0x3f3f3f3f;

 struct Ip

 {

     int a, b, c, d;

 }ip[MAXN];

 struct H

 {

     int a, b, c, d;

 }h[MAXN];

 int ans[];

 bool cmp(H x, H y)

 {

     if (x.a == y.a)

     {

         if (x.b == y.b)

         {

             if (x.c == y.c)    return x.d < y.d;

             else    return x.c < y.c;

         }

         else    return x.b < y.b;

     }

     else    return x.a < y.a;

 }

 int main(void)        //百度之星资格赛 1003 IP聚合

 {

     int t, cas = ;

     scanf ("%d", &t);

     while (t--)

     {

         int n, m;    scanf ("%d%d", &n, &m);

         for (int i=; i<=n; ++i)

             scanf ("%d.%d.%d.%d", &ip[i].a, &ip[i].b, &ip[i].c, &ip[i].d);

         memset (ans, , sizeof (ans));

         for (int i=; i<=m; ++i)

         {

             int u, v, w, x;

             scanf ("%d.%d.%d.%d", &u, &v, &w, &x);

             for (int j=; j<=n; ++j)

             {

                 h[j].a = (ip[j].a & u);    h[j].b = (ip[j].b & v);

                 h[j].c = (ip[j].c & w);    h[j].d = (ip[j].d & x);

             }

             sort (h+, h++n, cmp);

             int cnt = ;

             for (int k=; k<=n; ++k)

             {

                 if (h[k].a == h[k-].a &&

                     h[k].b == h[k-].b &&

                     h[k].c == h[k-].c &&

                     h[k].d == h[k-].d)    cnt++;

             }

             ans[i] = (n - cnt);

         }

         printf ("Case #%d:\n", ++cas);

         for (int i=; i<=m; ++i)    printf ("%d\n", ans[i]);

     }

     return ;

 }

 /*

 2

 5 2

 192.168.1.0

 192.168.1.101

 192.168.2.5

 192.168.2.7

 202.14.27.235

 255.255.255.0

 255.255.0.0

 4 2

 127.127.0.1

 10.134.52.0

 127.0.10.1

 10.134.0.2

 235.235.0.0

 1.57.16.0

 */