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: 
  import java.util.Scanner;
  
  public class Main {
  
      public static void main(String[] args) {
  
          int[][] input = getInput();
          for (int[] in : input) {
              computeAndPrintBrute(in);
          }
  
      }
  
      public static int[][] getInput() {
          Scanner scan = new Scanner(System.in);
          int num = scan.nextInt();
          int[][] input = new int[num][2];
  
  
          for (int i = 0; i < num; i++) {
              input[i][0] = scan.nextInt();
              input[i][1] = scan.nextInt();
          }
  
          return input;
      }
  
      public static void computeAndPrintBrute(int[] in) {
          System.out.println(new String(beautiful(in[0], in[1])));
      }
  
      // returns the position from the left that is not beautiful
      public static char[] beautiful(int in, int k) {
  
          char[] num = String.valueOf(in).toCharArray();
          boolean[] usedNum = new boolean[10];
          int used = 0;
  
          for (int i = 0; i < num.length; i++) {
              if (!usedNum[num[i] - 48]) {
                  if (++used > k) {
                      int mod = (int)Math.pow(10, num.length - i - 1);
                      in += mod;
                      in -= in % mod;
  
                      return beautiful(in, k);
                  }
                  usedNum[num[i] - 48] = true;
              }
          }
          return num;
      }
  }
  