別掉入 n 有 100000 的陷阱，實在太多了
利用他只有 10 個字版的特性去做所有的排列組合
最多只有 1024 個，並且利用排序的方式來參照字典有無這個字
速度會快很多
//====================================================================||
// ||
// ||
// Author : GCA ||
// 6AE7EE02212D47DAD26C32C0FE829006 ||
//====================================================================||
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <string\>
#include <climits>
#include <vector>
#include <set>
#include <map>
#include <queue>
#include <cctype>
#include <utility>
using namespace std;
#ifdef ONLINE\_JUDGE
#define ll "%lld"
#else
#define ll "%I64d"
#endif
typedef unsigned int uint;
typedef long long int Int;
#define Set(a,s) memset(a,s,sizeof(a))
#define Write(w) freopen(w,"w",stdout)
#define Read(r) freopen(r,"r",stdin)
#define Pln() printf("\\n")
#define I\_de(x,n)for(int i=0;i<n;i++)printf("%d ",x\[i\]);Pln()
#define De(x)printf(#x"%d\\n",x)
#define For(i,x)for(int i=0;i<x;i++)
#define CON(x,y) x##y
#define Pmz(dp,nx,ny)for(int hty=0;hty<ny;hty++){for(int htx=0;htx<nx;htx++){\\
printf("%d ",dp\[htx\]\[hty\]);}Pln();}
#define M 55
#define PII pair<int,int\>
#define PB push\_back
#define oo INT\_MAX
#define Set\_oo 0x3f
#define Is\_debug true
#define debug(...) if(Is\_debug)printf("DEBUG: "),printf(\_\_VA\_ARGS\_\_)
#define FOR(it,c) for(\_\_typeof((c).begin()) it=(c).begin();it!=(c).end();it++)
#define eps 1e-6
bool xdy(double x,double y){return x>y+eps;}
bool xddy(double x,double y){return x>y-eps;}
bool xcy(double x,double y){return x<y-eps;}
bool xcdy(double x,double y){return x<y+eps;}
int min3(int x,int y,int z){
int tmp=min(x,y);
return min(tmp,z);
}
int max3(int x,int y,int z){
int tmp=max(x,y);
return max(tmp,z);
}
int n,m,p;
set<string\> sw;
int wtile\[20\];
char ctile\[20\];
bool cmp(char a,char b){
return a<b;
}
int main() {
ios\_base::sync\_with\_stdio(0);
while(~scanf("%d",&n)){
for(int i=0;i<n;i++){
char str\[1000\];
scanf("%s",str);
int len=strlen(str);
sort(str,str+len,cmp);
sw.insert(str);
}
int nc;
scanf("%d",&nc);
while(nc--){
scanf("%d%\*c",&p);
for(int i=0;i<p;i++){
scanf("%c%d%\*c",&ctile\[i\],&wtile\[i\]);
}
int ans=0;
for(int i=0;i<(1<<p);i++){
string str;
int t=0;
for(int j=0;j<p;j++){
if((i>>j)&1){
str.PB(ctile\[j\]);
t+=wtile\[j\];
}
}
sort(str.begin(),str.end());
if(sw.count(str)){
ans=max(ans,t);
}
}
printf("%d\\n",ans);
}
}
}