description: 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.
solution: //package currentContest;
  
  import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.ArrayDeque;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.HashMap;
  import java.util.Random;
  import java.util.StringTokenizer;
  
  
  public class P1 {
  	public static void main(String[] args) {
  		// TODO Auto-generated method stub
  		FastReader sc=new FastReader();
  		int t=sc.nextInt();
  		StringBuilder st=new StringBuilder();
  		while(t--!=0){
  			String s=sc.nextLine();
  			String an=sc.nextLine();
  			char cs[]=s.toCharArray();
  			sort(cs);
  			if(cs[0]!='a'||an.charAt(0)!='a'||an.charAt(1)=='c') {
  				for(int i=0;i<s.length();i++) {
  					st.append(cs[i]);
  				}
  			}else {
  				int countb=0;
  				int flag=0;
  				for(int i=0;i<s.length();i++) {
  					if(cs[i]=='b') {
  						countb++;
  					}else if(cs[i]=='a') {
  						st.append(cs[i]);
  					}else if(cs[i]=='c'){
  						st.append(cs[i]);
  					}else {
  						while(countb!=0) {
  							st.append('b');
  							countb--;
  						}
  						st.append(cs[i]);
  					}
  					
  				}
  				while(countb!=0) {
  					st.append('b');
  					countb--;
  				}
  			}
  			st.append("\n");
  		}
  		System.out.println(st);
  	}
  	
  	static class DisjointUnionSets {
  		int[] rank, parent;
  		int n;
  		
  		// Constructor
  		public DisjointUnionSets(int n)
  		{
  			rank = new int[n];
  			parent = new int[n];
  			this.n = n;
  			makeSet();
  		}
  		
  		// Creates n sets with single item in each
  		void makeSet()
  		{
  			for (int i = 0; i < n; i++) {
  				// Initially, all elements are in
  				// their own set.
  				parent[i] = i;
  			}
  		}
  		
  		// Returns representative of x's set
  		int find(int x)
  		{
  			// Finds the representative of the set
  			// that x is an element of
  			if (parent[x] != x) {
  				// if x is not the parent of itself
  				// Then x is not the representative of
  				// his set,
  				parent[x] = find(parent[x]);
  				
  				// so we recursively call Find on its parent
  				// and move i's node directly under the
  				// representative of this set
  			}
  			
  			return parent[x];
  		}
  		
  		// Unites the set that includes x and the set
  		// that includes x
  		void union(int x, int y)
  		{
  			// Find representatives of two sets
  			int xRoot = find(x), yRoot = find(y);
  			
  			// Elements are in the same set, no need
  			// to unite anything.
  			if (xRoot == yRoot)
  				return;
  			
  			// If x's rank is less than y's rank
  			if (rank[xRoot] < rank[yRoot])
  				
  				// Then move x under y  so that depth
  				// of tree remains less
  				parent[xRoot] = yRoot;
  			
  			// Else if y's rank is less than x's rank
  			else if (rank[yRoot] < rank[xRoot])
  				
  				// Then move y under x so that depth of
  				// tree remains less
  				parent[yRoot] = xRoot;
  			
  			else // if ranks are the same
  			{
  				// Then move y under x (doesn't matter
  				// which one goes where)
  				parent[yRoot] = xRoot;
  				
  				// And increment the result tree's
  				// rank by 1
  				rank[xRoot] = rank[xRoot] + 1;
  			}
  		}
  	}
  
  	static FastReader sc = new FastReader();
  
  	public static void solvegraph() {
  		int n = sc.nextInt();
  
  		int edge[][] = new int[n - 1][2];
  		for (int i = 0; i < n - 1; i++) {
  			edge[i][0] = sc.nextInt() - 1;
  			edge[i][1] = sc.nextInt() - 1;
  		}
  		ArrayList<ArrayList<Integer>> ad = new ArrayList<>();
  		for (int i = 0; i < n; i++) {
  			ad.add(new ArrayList<Integer>());
  		}
  		for (int i = 0; i < n - 1; i++) {
  			ad.get(edge[i][0]).add(edge[i][1]);
  			ad.get(edge[i][1]).add(edge[i][0]);
  		}
  		int parent[] = new int[n];
  		Arrays.fill(parent, -1);
  		parent[0] = n;
  		ArrayDeque<Integer> queue = new ArrayDeque<>();
  		queue.add(0);
  		int child[] = new int[n];
  		Arrays.fill(child, 0);
  		ArrayList<Integer> lv = new ArrayList<Integer>();
  		while (!queue.isEmpty()) {
  			int toget = queue.getFirst();
  			queue.removeFirst();
  			child[toget] = ad.get(toget).size() - 1;
  			for (int i = 0; i < ad.get(toget).size(); i++) {
  				if (parent[ad.get(toget).get(i)] == -1) {
  					parent[ad.get(toget).get(i)] = toget;
  					queue.addLast(ad.get(toget).get(i));
  				}
  			}
  			lv.add(toget);
  		}
  
  		child[0]++;
  	}
  
  	static void sort(int[] A) {
  		int n = A.length;
  		Random rnd = new Random();
  		for (int i = 0; i < n; ++i) {
  			int tmp = A[i];
  			int randomPos = i + rnd.nextInt(n - i);
  			A[i] = A[randomPos];
  			A[randomPos] = tmp;
  		}
  		Arrays.sort(A);
  	}
  	static void sort(char[] A) {
  		int n = A.length;
  		Random rnd = new Random();
  		for (int i = 0; i < n; ++i) {
  			char tmp = A[i];
  			int randomPos = i + rnd.nextInt(n - i);
  			A[i] = A[randomPos];
  			A[randomPos] = tmp;
  		}
  		Arrays.sort(A);
  	}
  
  	static void sort(long[] A) {
  		int n = A.length;
  		Random rnd = new Random();
  		for (int i = 0; i < n; ++i) {
  			long tmp = A[i];
  			int randomPos = i + rnd.nextInt(n - i);
  			A[i] = A[randomPos];
  			A[randomPos] = tmp;
  		}
  		Arrays.sort(A);
  	}
  
  	static String sort(String s) {
  		Character ch[] = new Character[s.length()];
  		for (int i = 0; i < s.length(); i++) {
  			ch[i] = s.charAt(i);
  		}
  		Arrays.sort(ch);
  		StringBuffer st = new StringBuffer("");
  		for (int i = 0; i < s.length(); i++) {
  			st.append(ch[i]);
  		}
  		return st.toString();
  	}
  
  	public static long gcd(long a, long b) {
  		if (a == 0) {
  			return b;
  		}
  		return gcd(b % a, a);
  	}
  
  	static class FastReader {
  		BufferedReader br;
  		StringTokenizer st;
  
  		public FastReader() {
  			br = new BufferedReader(new InputStreamReader(System.in));
  		}
  
  		String next() {
  			while (st == null || !st.hasMoreElements()) {
  				try {
  					st = new StringTokenizer(br.readLine());
  				} catch (IOException e) {
  					e.printStackTrace();
  				}
  			}
  			return st.nextToken();
  		}
  
  		int nextInt() {
  			return Integer.parseInt(next());
  		}
  
  		long nextLong() {
  			return Long.parseLong(next());
  		}
  
  		double nextDouble() {
  			return Double.parseDouble(next());
  		}
  
  		String nextLine() {
  			String str = "";
  			try {
  				str = br.readLine();
  			} catch (IOException e) {
  				e.printStackTrace();
  			}
  			return str;
  		}
  	}
  }
  