#pragma comment(linker, "/STACK:102400000, 102400000")

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cctype>

#include<cmath>

#include<iostream>

#include<sstream>

#include<iterator>

#include<algorithm>

#include<string>

#include<vector>

#include<set>

#include<map>

#include<stack>

#include<deque>

#include<queue>

#include<list>

#define Min(a, b) ((a < b) ? a : b)

#define Max(a, b) ((a < b) ? b : a)

typedef long long ll;

typedef unsigned long long llu;

const int INT_INF = 0x3f3f3f3f;

const int INT_M_INF = 0x7f7f7f7f;

const ll LL_INF = 0x3f3f3f3f3f3f3f3f;

const ll LL_M_INF = 0x7f7f7f7f7f7f7f7f;

const int dr[] = {, , -, , -, -, , };

const int dc[] = {-, , , , -, , -, };

const int MOD = 1e9 + ;

const double pi = acos(-1.0);

const double eps = 1e-;

const int MAXN =  + ;

const int MAXT =  + ;

using namespace std;

char s[MAXN];

char t[MAXN];

vector<int> order;

vector<char> v[];

int POW[];

void init(){//2的i次方

    int tmp = ;

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

        tmp *= ;

        POW[i] = tmp;

    }

}

int main(){

    int n;

    int kase = ;

    init();

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

        if(n == ) return ;

        for(int i = ; i < ; ++i) v[i].clear();

        order.clear();

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

            char tmp[];

            scanf("%s", tmp);

            order.push_back(tmp[] - '');//S树的变量顺序

        }

        scanf("%s", s);

        int m;

        scanf("%d", &m);

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

            scanf("%s", t);

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

                v[i + ].push_back(t[i] - '');//存x[i+1]的m次查询的值

            }

        }

        printf("S-Tree #%d:\n", ++kase);

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

            int cnt = ;//根结点标号为1

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

                 if(!v[order[j]][i]){//若为0，则2i；否则2i+1

                    cnt *= ;

                 }

                 else{

                    cnt = cnt *  + ;

                 }

            }

            printf("%c", s[cnt - POW[n]]);

        }

        printf("\n\n");

    }

    return ;

}