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: for _ in range(int(input())):
      n, k = map(int, input().split())
      d = [int(x) for x in str(n)]
      l = len(d)
  
      if n == 1000000000:
          print(1111111111 if k == 1 else 1000000000)
          continue
      if k >= len(set(d)):
          print(n)
          continue
  
      mn = 10 ** 15
      for i in range(l):
          for v in range(d[i]+1, 10):
              uniq = set(d[:i]) | {v}
              if len(uniq) > k:
                  continue
              elif len(uniq) < k:
                  uniq |= {0}
              nd = d[:i] + [v] + [min(uniq)] * (l - i - 1)
              nn = int(''.join(map(str, nd)))
              if nn < mn:
                  mn = nn
      print(mn)
  