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 bisect
  import copy
  import decimal
  import fractions
  import functools
  import heapq
  import itertools
  import math
  import random
  import sys
  from collections import Counter,deque,defaultdict
  from functools import lru_cache,reduce
  from heapq import heappush,heappop,heapify,heappushpop,_heappop_max,_heapify_max
  def _heappush_max(heap,item):
      heap.append(item)
      heapq._siftdown_max(heap, 0, len(heap)-1)
  def _heappushpop_max(heap, item):
      if heap and item < heap[0]:
          item, heap[0] = heap[0], item
          heapq._siftup_max(heap, 0)
      return item
  from math import degrees, gcd as GCD
  read=sys.stdin.read
  readline=sys.stdin.readline
  readlines=sys.stdin.readlines
   
  T=int(readline())
  for _ in range(T):
      N,K=map(int,readline().split())
      S=str(N)
      l=len(S)
      ans=""
      se=set()
      m=-1
      for digit in range(l):
          for i in range(0,10):
              if digit==0 and i==0:
                  continue
              if int(S[:digit+1])>int(ans+str(i)):
                  continue
              if len(se)+(not i in se)<K:
                  ans+=str(i)
                  se.add(i)
                  m=max(m,i)
                  break
              elif len(se)+(not i in se)>K:
                  continue
              else:
                  if N<=int(ans+str(i)+str(max(m,i))*(l-digit-1)):
                      ans+=str(i)
                      se.add(i)
                      m=max(m,i)
                      break
      print(ans)