//UVA10817HeadmastersHeadache
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<sstream>
#include<string>
using namespace std;
const int MAXS = 8;
const int MAXN = 120 + 5;
const int INF = 1e9;
int s, m, n;
int cost[MAXN], st[MAXN], d[MAXN][1 << MAXS][1 << MAXS];
int dp(int i, int s0, int s1, int s2) {
//printf("****, i = %d, m = %d, n = %d\n", i, m, n);
if(i == m + n) return s2 == (1 << s) - 1 ? 0 : INF;
int& ans = d[i][s1][s2];
if(ans >= 0) return ans; //记忆化,避免重复
ans = INF;
if(i >= m) ans = dp(i + 1, s0, s1, s2); //此时的决策有两种，先考虑一种
int m0 = st[i] & s0, m1 = st[i] & s1;
s0 ^= m0;//m0中的元素即将被添加到s1中，将s0中的m0部分除去
s1 = (s1 ^ m1) | m0;//m1将被加入s2，所以要从s1中除去
s2 |= m1;
ans = min(ans, dp(i + 1, s0, s1, s2) + cost[i]);
return ans;
}
int main() {
string line;
while(getline(cin, line)) {
memset(d, -1, sizeof(d));
memset(st, 0, sizeof(st));
stringstream ss(line);
ss >> s >> m >> n;
if(s == 0) return 0;
for(int i = 0; i < m + n; i++) {
getline(cin, line);
stringstream s1(line);
s1 >> cost[i];
int x;
while(s1 >> x) st[i] |= (1 << (x - 1));
}
printf("%d\n", dp(0, (1 << s) - 1, 0, 0));
}
return 0;
}
/*
2 2 2
10000 1
20000 2
30000 1 2
40000 1 2
0 0 0
*/
