/*

      题意：给出一个数，问是否有ai + bj + ck == x

      二分查找：首先计算sum[l] = a[i] + b[j]，对于q，枚举ck，查找是否有sum + ck == x

 */

 #include <cstdio>

 #include <algorithm>

 #include <cmath>

 using namespace std;

 typedef long long ll;

 const int MAXN = 5e2 + ;

 const int INF = 0x3f3f3f3f;

 ll a[MAXN], b[MAXN], c[MAXN];

 ll sum[MAXN*MAXN];

 int tot;

 bool my_binary_search(int l, int r, ll k)   {

     while (l < r)   {

         int mid = (l + r) >> ;

         if (sum[mid] == k)  return true;

         else if (sum[mid] > k)  r = mid;

         else    l = mid + ;

     }

     return false;

 }

 int main(void)  {       //HDOJ 2141 Can you find it?

     //freopen ("HDOJ_2141.in", "r", stdin);

     int l, n, m, s, cas = ;

     while (scanf ("%d%d%d", &l, &n, &m) == )   {

         for (int i=; i<=l; ++i)    scanf ("%I64d", &a[i]);

         for (int i=; i<=n; ++i)    scanf ("%I64d", &b[i]);

         for (int i=; i<=m; ++i)    scanf ("%I64d", &c[i]);

         scanf ("%d", &s);

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

         sort (a+, a++l);  sort (b+, b++n);  sort (c+, c++m);

         tot = ;

         for (int i=; i<=l; ++i)    {

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

                 sum[++tot] = a[i] + b[j];

             }

         }

         sort (sum+, sum++tot);

         ll mn = a[] + b[] + c[], mx = a[l] + b[n] + c[m];

         while (s--) {

             ll q;  scanf ("%I64d", &q);

             if (q < mn || q > mx)   {

                 puts ("NO");    continue;

             }

             bool flag = false;

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

                 int p = lower_bound (sum+, sum++tot, q - c[i]) - sum;

                 if (p <  || p > tot)   continue;

                 if (sum[p] + c[i] == q) {

                     flag = true;    puts ("YES");   break;

                 }

                 //if (my_binary_search (1, tot, q - c[i]))    {

                     //flag = true;    puts ("YES");   break;

                 //}

             }

             if (!flag)  {

                 puts ("NO");

             }

         }

     }

     return ;

 }