typedef unsigned long long ull;

const int N = 100000 + 5;

const ull base = 163;

char s[N];

ull hash[N];

 

void init(){//处理hash值

    p[0] = 1;

    hash[0] = 0;

    int n = strlen(s + 1);

   for(int i = 1; i <=100000; i ++)p[i] =p[i-1] * base;

   for(int i = 1; i <= n; i ++)hash[i] = hash[i - 1] * base + (s[i] - 'a');

}

 

ull get(int l, int r, ull g[]){//取出g里l - r里面的字符串的hash值

    return g[r] - g[l - 1] * p[r - l + 1];

}

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. 

Range of test data: 
T<= 100 ; 
n<= 100000; 

2

abcdefghijklmnopqrstuvwxyz

abcdab

qwertyuiopasdfghjklzxcvbnm

qwertabcde

abcdabcd

qwertabcde

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<iomanip>

#include<string>

#include<algorithm>

using namespace std;

#define inf 1<<29

#define nu 1000005

#define maxnum 100005

#define num 30

typedef unsigned long long ull;

int t,n;

int len;

ull m=161;

ull h1[maxnum],h2[maxnum],p[maxnum];

char s[30],ch[maxnum];

int c[30];

void getp(){

    p[0]=1;

    for(int i=1;i<=100005;i++)

        p[i]=p[i-1]*m;

}

ull geth(ull h[],int i,int j){

    return h[j]-h[i-1]*p[j-i+1];

}

void solved(){

    h1[0]=h2[0]=0;

    for(int i=0;i<26;i++)

        c[s[i]-'a']=i;

    for(int i=1;i<=len;i++){

        h1[i]=h1[i-1]*m+c[ch[i]-'a'];

         h2[i]=h2[i-1]*m+ch[i]-'a';

    }

    int a=len/2+len%2;

    int flag;

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

            int le=len-i;

        ull key1=geth(h1,1,le);

        ull key2=geth(h2,i+1,len);

        //cout<<key1<<" "<<key2<<endl;

        if(key1==key2)

        {

           flag=i;

           break;

        }

    }

   // cout<<flag;

for(int i=1;i<=flag;i++)

           printf("%c",ch[i]);

    for(int i=1;i<=flag;i++)

    {

        char cc=c[ch[i]-'a']+'a';

        printf("%c",cc);

    }

    puts("");

}

int main()

{

    scanf("%d",&t);

    getp();

    while(t--){

        scanf("%s%s",s,ch+1);

        len=strlen(ch+1);

        solved();

    }

    return 0;

}