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: def f(n,k):
      dig_set = set()
      pre = ''
      for x in n:
          dig_set.add(x)
          if len(dig_set)>int(k): 
              post_head = x
              dig_set.remove(x)
              break
          pre += x
          post_head=""
  
      post = n[len(pre):] 
  
      if post_head:
          min_dig = min(dig_set)
          poss_digs = {i for i in dig_set if i>x}
          if poss_digs:
              min_poss_dig = min(poss_digs)
              ans = pre+min_poss_dig+min_dig*(len(post)-1)
          else:
              pre = str(int(pre)+1)
              if len(set(pre))==int(k):
                  ans = pre+min(set(pre))*(len(post))
              elif len(set(pre))<int(k):
                  ans = pre+"0"*(len(post))
              else:
                  ans = f(pre+"0"*(len(post)),k)
      else:
          ans = pre
      
      return ans
  
  for i in range(int(input())):
      n,k = input().split()
      print(f(n,k))