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 upList(s, i):
      left = s[:i]
      right = s[i:]
      left = list(str(int(''.join(left)) + 1))
      return left + right
  
  def jg(x, val):
      y = set(i for i in x if i >= val)
      if len(y) == 0:
          return None
      return min(y)
  
  
  for _ in range(int(input())):
      n, k = map(int, input().split())
      s = list(str(n))
  
      i = len(s) - 1
      while i >= 0:
          cMax = set()
          j = 0
          while len(cMax) < k and j < len(s):
              cMax.add(s[j])
              j += 1
          if j == len(s):
              break
  
          if s[i] not in cMax:
              z = jg(cMax, s[i])
              if z == None:
                  s = upList(s, i)
                  for var in range(i, len(s)):
                      s[var] = '0'
                  i = len(s) - 1
              else:
                  s[i] = z
                  for var in range(i + 1, len(s)):
                      s[var] = min(cMax)
                  i -= 1
          else:
              i -= 1
          
      print(''.join(s))