/*

     题意：学校有在任的老师和应聘的老师，选择一些应聘老师，使得每门科目至少两个老师教，问最少花费多少

     状压DP：一看到数据那么小，肯定是状压了。这个状态不好想，dp[s1][s2]表示s1二进制表示下至少有1位老师的科目集合

             s2表示至少有2位老师的科目集合所花费的最小金额，状态转移方程(01)：dp[t1][t2]=min(dp[t1][t2],dp[j][k]+c[i]);

             j,k为当前两个集合，t1，t2为转移后的集合，另外求t1，t2用到了& |位运算 1&1 == 1 1 & 0 == 0 0 & 0 == 0

             最后，学习了不知道数字多少个时应该用字符串整行读入

 　　详细解释

 */

 /************************************************

 * Author        :Running_Time

 * Created Time  :2015-8-10 14:44:30

 * File Name     :UVA_10817.cpp

  ************************************************/

 #include <cstdio>

 #include <algorithm>

 #include <iostream>

 #include <sstream>

 #include <cstring>

 #include <cmath>

 #include <string>

 #include <vector>

 #include <queue>

 #include <deque>

 #include <stack>

 #include <list>

 #include <map>

 #include <set>

 #include <bitset>

 #include <cstdlib>

 #include <ctime>

 using namespace std;

 #define lson l, mid, rt << 1

 #define rson mid + 1, r, rt << 1 | 1

 typedef long long ll;

 const int MAXN = ;

 const int INF = 0x3f3f3f3f;

 const int MOD = 1e9 + ;

 int c[MAXN], p[MAXN], cnt[];

 int dp[(<<)+][(<<)+];

 int s, m, n;

 int mxs;

 int sum, st1, st2;

 int work(void)  {

     memset (dp, INF, sizeof (dp));

     dp[st1][st2] = sum;

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

         for (int j=mxs; j>=; --j) {

             for (int k=mxs; k>=; --k) {

                 if (dp[j][k] == INF)    continue;

                 int t1 = (p[i] | j);  int t2 = (p[i] & j) | k;

                 dp[t1][t2] = min (dp[t1][t2], dp[j][k] + c[i]);

             }

         }

     }

     return dp[mxs][mxs];

 }

 int main(void)    {     //UVA 10817 Headmaster's Headache

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

         if (!s) break;

         memset (cnt, , sizeof (cnt));

         memset (p, , sizeof (p));

         sum = , st1 = st2 = ; mxs = ( << s) - ;

         string str;

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

             scanf ("%d", &c[i]);

             getline (cin, str);

             for (int j=; str[j]; ++j)  {

                 if (isdigit (str[j]))   {

                     int x = str[j] - '';

                     p[i] |=  << (x - );

                     if (i <= m) ++cnt[x-];

                 }

             }

             if (i <= m) {

                 sum += c[i];    st1 |= p[i];

             }

         }

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

             if (cnt[i] > ) st2 |= ( << i);

         }

         printf ("%d\n", work ());

     }

     return ;

 }