Description

Clairewd is a member of FBI. After several years concealing in BUPT, she intercepted some important messages and she was PReparing for sending it to ykwd. They had agreed that each letter of these messages would be transfered to another one according to a conversion table.

    Unfortunately, GFW(someone's name, not what you just think about) has detected their action. He also got their conversion table by some unknown methods before. Clairewd was so clever and vigilant that when she realized that somebody was monitoring their action, she just stopped transmitting messages.

    But GFW knows that Clairewd would always firstly send the ciphertext and then plaintext(Note that they won't overlap each other). But he doesn't know how to separate the text because he has no idea about the whole message. However, he thinks that recovering the shortest possible text is not a hard task for you.

    Now GFW will give you the intercepted text and the conversion table. You should help him work out this problem.

Input

The first line contains only one integer T, which is the number of test cases.

Each test case contains two lines. The first line of each test case is the conversion table S. S[i] is the ith latin letter's cryptographic letter. The second line is the intercepted text which has n letters that you should recover. It is possible that the text is complete.

T<= 100 ;n<= 100000;

Output

For each test case, output one line contains the shorest possible complete text.

Sample Input

2
abcdefghijklmnopqrstuvwxyz
abcdab
qwertyuiopasdfghjklzxcvbnm
qwertabcde
Sample Output
abcdabcdqwertabcde
HINT
Source
介個題目的意思就是給你一個26個字母的對照表，就是例如qwertyuiopasdfghjklzxcvbnm 表示q對應的a，w對應的b，e對應的c.....類推m對應的z。然后再給你一個字符串，密碼包括密文+明文（這個概念我理解了好久-。-）qwert就是密文，然后qwert按對照表出來的明文就是abcde，然鵝，這個字符串可能是殘缺的，它并不一定是密碼！只有密文是一定全部都在的。所以就要從中間開始找啦。你要做的就是找到密文和明文的分界點，輸出密文+明文。具體怎么找 ↓↓↓
#include <stdio.h>#include <string.h>int main(){    int n;    scanf("%d",&n);    while(n--)    {        char a[100]={0},b[100101]={0},dic[200]={0};        scanf("%s%s",a,b);        int n=strlen(b),m,w=97;        m=(1+n)/2;   //一開始用了n/2 結果一直錯，后來發現有點小問題-0-  n為奇數的時候可能會出現中間字母恰好和第一個和最后一個對應                     //例一來說 abbabba 輸出本應該是abbabbabbabb  但是n/2 就會是abbabb        int i=0,j;        for(j=0;j<strlen(a);j++)dic[a[j]]=w++;        while((m+i)<n)        {            if(dic[b[i]]==b[i+m])i++;            else {m++;i=0;}        }            for(i=0;i<m;i++)printf("%c",b[i]);            for(i=0;i<m;i++)printf("%c",dic[b[i]]);            printf("/n");    }    return 0;}