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.
#Mamma don't raises quitter.................................................
#manan is improving, keep hustling manan bhai ;)
from collections import deque as de
import math
import sys
from math import sqrt as sq
from math import floor as fl
from math import ceil as ce
from sys import stdin, stdout

sys.setrecursionlimit(10**7)
from collections import Counter as cnt
from functools import  reduce

from itertools import groupby as gb
#from fractions import Fraction as fr
from bisect import bisect_left as bl, bisect_right as br
from typing import Coroutine

def factors(n):    
    return set(reduce(list.__add__, 
                ([i, n//i] for i in range(1, int(n**0.5) + 1) if n % i == 0)))
    
class My_stack():
    def __init__(self):
        self.data = []
    def my_push(self, x):
        return (self.data.append(x))
    def my_pop(self):
        return (self.data.pop())
    def __iter__(self):
        return iter(self.data)
    def my_peak(self):
        return (self.data[-1])
    def my_contains(self, x):
        return (self.data.count(x))
    def my_show_all(self):
        return (self.data)
    def isEmpty(self):
      return len(self.data)==0
    def len(self):
        return len(self.data)
    

arrStack = My_stack() 

class graph():
    def __init__(self):
        self.graphdict = {}
    def add_edge(self, node, neighbor):
        if node not in self.graphdict:
            self.graphdict[node]=[neighbor]
        else:
            self.graphdict[node].append(neighbor)
    def show_edge(self):
        for node in self.graphdict:
            for neighbors in self.graphdict:
                print(node, neighbors)
    def show_graph(self):
        return self.graphdict

    


    
            

#decimal to binary   
def decimalToBinary(n): 
    return bin(n).replace("0b", "")
#binary to decimal
def binarytodecimal(n):
    return int(n,2)

def isPrime(n) : 
	if (n <= 1) : 
		return False
	if (n <= 3) : 
		return True
	if (n % 2 == 0 or n % 3 == 0) : 
		return False
	i = 5
	while(i * i <= n) : 
		if (n % i == 0 or n % (i + 2) == 0) : 
			return False
		i = i + 6
	return True

def get_prime_factors(number):
    prime_factors = []
    while number % 2 == 0:
        prime_factors.append(2)
        number = number / 2
    for i in range(3, int(math.sqrt(number)) + 1, 2):
        while number % i == 0:
            prime_factors.append(int(i))
            number = number / i

    if number > 2:
        prime_factors.append(int(number))
    return prime_factors

def get_frequency(list):
    dic={}
    for ele in list:
        if ele in dic:
            dic[ele] += 1
        else:
            dic[ele] = 1
    return dic

def Log2(x): 
    return (math.log10(x) / 
            math.log10(2)); 

 
# Function to get product of digits
def getProduct(n):
    product = 1
    while (n != 0):
        product = product * (n % 10)
        n = n // 10
    return product
#fucntion to find regular bracket sequence
def regularBracketsequence(str,arrStack):
    for i in range(len(str)):
        if str[i] =='(':
            arrStack.my_push(str[i])
        else:
            if arrStack.isEmpty():
                return False
            else:
                arrStack.my_pop()
    if arrStack.isEmpty():
        return True
    else:
        return False

#function to find LCM of two numbers
def lcm(x,y):
   lcm = (x*y)//math.gcd(x,y)
   return lcm

def isPowerOfTwo(n): 
    return (math.ceil(Log2(n)) == math.floor(Log2(n))); 
#to check whether the given sorted sequnce is forming an AP or not....
def checkisap(list):
    d=list[1]-list[0]
    for i in range(2,len(list)):
        temp=list[i]-list[i-1]
        if temp !=d:
            return False
    return True

#seive of erathanos
def primes_method5(n):
    out ={}
    sieve = [True] * (n+1)
    for p in range(2, n+1):
        if (sieve[p]):
            out[p]=1
            for i in range(p, n+1, p):
                sieve[i] = False
    return out
#fucntion to get unique number to a cell in a matrix
def id(i, j, col): #one based indexing and columnwise numeration
    ans= (i-1)*col
    ans+=j
    return ans

#function to get the sum of digits
def getSum(n): 
      
    strr = str(n)
    list_of_number = list(map(int, strr.strip()))
    return sum(list_of_number)
def checkperfectsquare(n):
    if (ce(sq(n)) == fl(sq(n))):
        return True
    else:
        return False
#this used for selecting r objects from n ojects
def combinatrics(n,r):
    num=n
    deno=n-r
    pr=1
    for i in range(num,deno,-1):
        pr*=i
    an=1
    for i in range(1,r+1):
        an*=i
    ans=pr//an
    return ans
#ceil  function gives wrong answer after 10^17 so i have to create my own :)
# because i don't want to doubt on my solution of 900-1000 problem set.
def ceildiv(x,y): 
    return (x+y-1)//y 
  
def di():return map(int, input().split())
def ii():return int(input())
def li():return list(map(int, input().split()))
def si():return list(map(str, input()))
def indict():
    dic = {}
    for index, value in enumerate(input().split()):
        dic[int(value)] = int(index)
    return dic
def frqdict(): 
    # by default it is for integer input. :)
    dic={}
    for index, value in enumerate(input()):
        if value not in dic:
            dic[value] =1
        else:
            dic[value] +=1
    return dic

#sys.stdin = open('input.txt', 'r')
#sys.stdout = open('output.txt', 'w')
#practice like your never won
#perform like you never lost
t=ii()
while t:
    t-=1
    s=si()
    ts=input()
    if ts=="abc":
        ac=s.count("a")
        bc=s.count("b")
        cc=s.count("c")
        s.sort()
        if ac and bc and cc:
            temp="".join(s).rfind("c")
            final="a"*(ac)+"c"*(cc)+"b"*(bc)+"".join((s[temp+1:]))
            print(final)
        else:
            s.sort()
            print("".join(s))
            

    else:
        s.sort()
        print("".join(s))

    


        
        