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 solve():
      n, k = map(int, input().split())
      s = str(n)
      se = set()
      for i in range(len(s)):
          se.add(s[i])
      if len(se) <= k:
          return s
      se = set()
      i = 0
      while True:
          se.add(s[i])
          if len(se) > k:
              break
          i += 1
      se.remove(s[i])
      arr = sorted(se)
      while True:
          cur = set(c for c in s[:i])
          if len(cur) == k:
              if s[i] >= arr[-1]:
                  i -= 1
                  continue
              j = 0
              while arr[j] <= s[i]:
                  j += 1
              return s[:i] + arr[j] + arr[0]*(len(s)-i-1)
          c = str(int(s[i])+1)
          cur.add(c)
          r = min(cur) if len(cur) == k else "0"
          return s[:i] + c + r*(len(s)-i-1)
  
  
  
  import sys
  input = lambda: sys.stdin.readline().rstrip()
  t = int(input())
  for i in range(t):
      print(solve())