You are given strings S and T, consisting of lowercase English letters. It is guaranteed that T is a permutation of the string abc. 

Find string S', the lexicographically smallest permutation of S such that T is not a subsequence of S'.

String a is a permutation of string b if the number of occurrences of each distinct character is the same in both strings.

A string a is a subsequence of a string b if a can be obtained from b by deletion of several (possibly, zero or all) elements.

A string a is lexicographically smaller than a string b if and only if one of the following holds:

  * a is a prefix of b, but a ≠ b;
  * in the first position where a and b differ, the string a has a letter that appears earlier in the alphabet than the corresponding letter in b.

Input

Each test contains multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases. Description of the test cases follows.

The first line of each test case contains a string S (1 ≤ |S| ≤ 100), consisting of lowercase English letters.

The second line of each test case contains a string T that is a permutation of the string abc. (Hence, |T| = 3).

Note that there is no limit on the sum of |S| across all test cases.

Output

For each test case, output a single string S', the lexicographically smallest permutation of S such that T is not a subsequence of S'.

Example

Input


7
abacaba
abc
cccba
acb
dbsic
bac
abracadabra
abc
dddddddddddd
cba
bbc
abc
ac
abc


Output


aaaacbb
abccc
bcdis
aaaaacbbdrr
dddddddddddd
bbc
ac

Note

In the first test case, both aaaabbc and aaaabcb are lexicographically smaller than aaaacbb, but they contain abc as a subsequence.

In the second test case, abccc is the smallest permutation of cccba and does not contain acb as a subsequence.

In the third test case, bcdis is the smallest permutation of dbsic and does not contain bac as a subsequence.
# cook your dish here
# cook your dish here
#-------------------imports-------#
import sys
import io,os
import string
from io import BytesIO, IOBase
from os import path
from collections import Counter,OrderedDict
import itertools

from itertools import chain, combinations
import math
import random
import bisect
from functools import reduce
from sys import stdin,stdout
from math import gcd,floor,sqrt,log
from collections import defaultdict as dd
from bisect import bisect_left as bl,bisect_right as br


#------------------ONLINE JUDGE OR NOT-------#

if(path.exists('input.txt')):
    sys.stdin = open("input.txt","r")
    sys.stdout = open("output.txt","w")
else:
    input = sys.stdin.readline
    # input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline



sys.setrecursionlimit(100000000)



#---------------------pre made and input functions-------#
simp    =lambda: int(input())
strng  =lambda: input().strip()
jn     =lambda x,l: x.join(map(str,l))
strl   =lambda: list(input().strip())
mul    =lambda: map(int,input().strip().split())
mulf   =lambda: map(float,input().strip().split())
seq    =lambda: list(map(int,input().strip().split()))
ceil   =lambda x: int(x) if(x==int(x)) else int(x)+1
ceildiv=lambda x,d: x//d if(x%d==0) else x//d+1
flush  =lambda: stdout.flush()
stdstr =lambda: stdin.readline()
stdint =lambda: int(stdin.readline())
stdpr  =lambda x: stdout.write(str(x))


#------------functions---------------------

#---------------------input-------

a = simp()


def if_its_binaryform_palindrome(n):

    res = 0

    k = n
    while k > 0:
        #to finish the job
        res = (res << 1) | (k & 1)
        k = k >> 1  

    return n == res

# prime = list_of_prime_sieve(10000000)        
  
def solve(s,t):

    first = t[0]
    second = t[1]
    third = t[2]

    if 'a' not in s or 'b' not in s or 'c' not in s:
        return s


    if s[0]=='a' and t[0]!='a':
        sortedd = sorted(s)
        s = ''.join(sortedd)
        # s.sort()
        return s

    sortedd = sorted(s)
    s = ''.join(sortedd)    
    # s.sort()
        
    d = Counter(s)  

    r = ''
    if t[1]=='b':
        rr = ''
        for i,j in d.items():
            if i=='a':
                r+=i*j
            elif i=='b' and r[-1]=='a':
                rr+=i*j
            elif i=='c':
                r+=i*j
                r+=rr
            else:
                r+=i*j

        return r
    elif t[1]=='c':
        rr = ''
        for i,j in d.items():
            if i=='a':
                r+=i*j
            elif i=='c' and r[-1]=='a':
                rr+=i*j
            elif i=='b':
                r+=i*j
                r+=rr
            else:
                r+=i*j
        return r

    elif t[1]=='a':
        rr = ''
        for i,j in d.items():
            r+=i*j

    return r       

                        


                    
                    


    # a a a a b b c


    # abc

    # t = 'abc'    



for i in range(a):
    t = strng()
    d = strng()
    print(solve(t,d))















def seive(n):
    """ preprocessing  """
 
    prime = [1 for i in range(n + 1)]
    p = 2
    while (p * p <= n):
         
        # this is prime number.
        if (prime[p] == 1):
             
            # Update all multiples of p in array
            for i in range(p ** 2, n + 1, p):
                prime[i] = 0
                
        p += 1
    prime[0]= 0
    prime[1]= 0
    
    # print(prime)
    for i in range(1,n+1):
        prime[i] += prime[i-1]
        
    return prime

def list_of_prime_sieve(n):

    #an array of True True
    primes = [True]*(n+1)
    #changing o and 1 to False
    primes[0]==primes[1]==False

    #now for every true we find
    for i in range(2,int(n**(1/2))+1):
        if primes[i]==True:
            #we change their multiples to False
            for j in  range(i*i,n+1,i):
                primes[j]=False


    list_primes = []
    for i in range(2,len(primes)):
        if primes[i]==True:
            list_primes.append(i)

    return list_primes

def count_primes(a):
    return len(list_of_prime_sieve(a))    


def gcdoftwo(a,b):

    #log(min(a,b))
    if a == 0:
        return b
    else:
        return gcdoftwo(b%a,a)

def lcm(a,b): return (a*b//(gcdoftwo(a,b)))        

def gcdofList(A): 
    res = A[0]
    for c in A[1::]:
        res = gcdoftwo(res , c) 
        if res == 1:
            return res
    return res         

def binpow(a,b):

    #to calc a^b in log(n) time


    res = 1


    while b>0:

        if b&1:
            res = res*a
        a = a*a
        b>>=1
    return res        

def fibo_in_logn(n):
    if n == 0:
        return [0, 1]

    p = fibo_in_logn(n >> 1)
    c = p.first * (2 * p.second - p.first)
    d = p.first * p.first + p.second * p.second
    if n & 1:
        return {d, c + d}
    else:
        return {c, d}





def x_n_mod_m(a,b,m):

    #to calculate x^n mod m

    a  = a%m 

    res = 1
    while b>0:
        if b&1:
            res = res*a%m 
        a = a*a%m
        b>>=1

    return res




def change_array(count,t):

    emp = []
    for i in range(len(t)):
        emp.append(count[t[i]])
    return emp    




# prime = []

def simpleSieve(limit):

    mark = [True for i in range(limit + 1)]
    p = 2
    while (p * p <= limit):
        if (mark[p] == True):
            for i in range(p * p, limit + 1, p):
                mark[i] = False
        p += 1
    for p in range(2, limit):
        if mark[p]:
            prime.append(p)
            # print(p,end = " ")
    return prime        

def segmentedSieve(n):
    limit = int(math.floor(math.sqrt(n)) + 1)
    t = []
    t += simpleSieve(limit)

    low = limit
    high = limit * 2
    while low < n:
        if high >= n:
            high = n

        mark = [True for i in range(limit + 1)]

        for i in range(len(prime)):

            loLim = int(math.floor(low / prime[i]) *
                                        prime[i])
            if loLim < low:
                loLim += prime[i]
            for j in range(loLim, high, prime[i]):
                mark[j - low] = False

        # t = []        
        for i in range(low, high):
            if mark[i - low]:
                t.append(i)
                # print(i, end = " ")

        low = low + limit
        high = high + limit

    return t




mod=1000000007

def occurrences(string, sub):
    count = start = 0
    while True:
        start = string.find(sub, start) + 1
        if start > 0:
            count+=1
        else:
            return count
























    
















