description: It is a complicated version of problem F1. The difference between them is the constraints (F1: k ≤ 2, F2: k ≤ 10).
  
  You are given an integer n. Find the minimum integer x such that x ≥ n and the number x is k-beautiful.
  
  A number is called k-beautiful if its decimal representation having no leading zeroes contains no more than k different digits. E.g. if k = 2, the numbers 3434443, 55550, 777 and 21 are k-beautiful whereas the numbers 120, 445435 and 998244353 are not.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t test cases follow.
  
  Each test case consists of one line containing two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 10).
  
  Output
  
  For each test case output on a separate line x — the minimum k-beautiful integer such that x ≥ n.
  
  Example
  
  Input
  
  
  6
  2021 3
  177890 2
  34512 3
  724533 4
  998244353 1
  12345678 10
  
  
  Output
  
  
  2021
  181111
  34533
  724542
  999999999
  12345678
solution: //package training;
  import java.util.*;
  public class Training {
      public static void main(String args[]) { 
          Scanner in = new Scanner (System.in);
          int t = in.nextInt();
          for (int i = 0; i < t; i++) {
              int n = in.nextInt();
              int k = in.nextInt();
              char [] chars = String.valueOf(n).toCharArray();
              while (true){
                  Set<Character> s = new HashSet<>();
                  for (char c : chars) {
                      s.add(c);
                  }
                  if (s.size() <= k) {
                      break;
                  }
                  s.clear();
   
                  int ptr = 0;
                  for (; ; ptr++) {
                      s.add(chars[ptr]);
                      if (s.size() > k) {
                          while (chars[ptr] == '9') {
                              ptr--;
                          }
                          chars[ptr]++;
                          for (int j = ptr + 1; j < chars.length; j++) {
                              chars[j] = '0';
                          }
                          break;
                      }
                  }
              }
              System.out.println(new String(chars));
          }
      }
  }
  