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 sys
  import copy
  
  INF = 10**10
  
  def digs(n):
      ans = []
      while n > 0:
          ans.append(n % 10)
          n //= 10
      return list(reversed(ans))
  
  def find_ans(n, k):
      n_digs = digs(n)
      if k == 1:
          base = (10**len(n_digs)-1)//9
          for d in range(1, 10):
              if d*base >= n:
                  return d*base
              
          print("ANS NOT FOUND")
          sys.exit(1)
  
      ans = INF
      nums_found = set()
      bad = False
      prefix = 0
      for i, d in enumerate(n_digs):
          realLen = i
  
          next_dig = d+1
          while len(nums_found)+int(next_dig not in nums_found) > k:
              next_dig += 1
              if next_dig > 9:
                  break
              
          if next_dig <= 9:
              new_prefix = prefix*10+next_dig
              test_ans = new_prefix*10**(len(n_digs)-i-1)
  
              new_nums_found = copy.deepcopy(nums_found)
              new_nums_found.add(next_dig)
              if len(new_nums_found) == k:
                  test_ans += min(new_nums_found)*(10**(len(n_digs)-i-1)-1)//9
              ans = min(test_ans, ans)
          
          prefix *= 10
          prefix += d
          nums_found.add(d)
          
      if len(nums_found) <= k:
          return n
      return ans
  
  
  nums = (int(tok) for tok in sys.stdin.read().split())
  T = next(nums)
  for _ in range(T):
      n = next(nums)
      k = next(nums)
      
      print(find_ans(n, k))
  