Using regular expression to define a numeric string is a very common thing. Generally, use the shape as follows:

(||) (|) () (|)

Above regular expression matches  digits:The first is one of , and . The second is one of  and . The third is . And the fourth is one of  and . The above regular expression can be successfully matched to , but it cannot be matched to .

Now,giving you a regular expression like the above formula,and a long string of numbers,please find out all the substrings of this long string that can be matched to the regular expression.

Input

It contains a set of test data.The first line is a positive integer N ( ≤ N ≤ ),on behalf of the regular representation of the N bit string.In the next N lines,the first integer of the i-th line is ai(≤ai≤)ai(≤ai≤),representing that the i-th position of regular expression has aiai numbers to be selected.Next there are aiai numeric characters. In the last line,there is a numeric string.The length of the string is not more than  * ^.

Output

Output all substrings that can be matched by the regular expression. Each substring occupies one line

Sample Input

Sample Output

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<sstream>

#include<algorithm>

#include<queue>

#include<deque>

#include<iomanip>

#include<vector>

#include<cmath>

#include<map>

#include<stack>

#include<set>

#include<memory>

#include<list>

#include<bitset>

#include<string>

#include<functional>

using namespace std;

typedef long long LL;

typedef unsigned long long ULL;

const int MAXN = 5e6 + ;

#define L 1009

#define INF 1000000009

#define eps 0.00000001

#define MOD 1000

bitset<> B[], D;

char str[MAXN];

int main()

{

    int n, tmp, t;

    scanf("%d", &n);

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

    {

        scanf("%d", &tmp);

        while (tmp--)

        {

            scanf("%d", &t);

            B[t].set(i);

        }

    }

    getchar();

    gets(str);

    int l = strlen(str);

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

    {

        D = (D << ).set()&B[str[i] - ''];

        if (D[n - ])

        {

            char ch = str[i + ];

            str[i + ] = '\0';

            puts(str + i - n + );

            str[i + ] = ch;

        }

    }

}