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: import java.io.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.lang.reflect.Array;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.StringTokenizer;
  
  public class pre187 {
      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;
          }
      }
      public static void main(String args[]){
          FastReader obj = new FastReader();
          int tc = obj.nextInt();
          while(tc--!=0){
              char s[] = obj.next().toCharArray(),t[] = obj.next().toCharArray();
              Arrays.sort(s);
              ArrayList<Integer> mp[] = new ArrayList[3];
              boolean has = true;
              while(true){
                  int count = 0;
                  if(!has) break;
                  has = false;
                  ArrayList<Integer> set = new ArrayList<>();
                  for(int i=0;i<s.length;i++){
                      if(s[i]==t[count]){
                          count++;
                          set.add(i);
                      }
                      if(count==3) {
                          has = true;
                          break;
                      }
                  }
                  if(has){
                      char temp = s[set.get(1)];
                      s[set.get(1)] = s[set.get(2)];
                      s[set.get(2)] = temp;
                  }
              }
              System.out.println(String.valueOf(s));
          }
      }
  }
  