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 lower_bound(nums,x):
      for i in range(len(nums)):
          if nums[i]>=x:
              return i
      return -1
  
  def check(n,k,num,pos,_n,wid):
      x=num
      nums=dict()
      while x>0:
          nums.setdefault(x%10,0)
          x//=10
      nums=list(nums.keys())
      nums.sort()
      flag=0
      for i in range(pos,wid):
          tmp=int(_n[i])
          x=lower_bound(nums,tmp) if flag==0 else 0
          if x==-1:
              return -1
          if flag==0 and nums[x]>tmp:
              flag=1
          num=num*10+nums[x]
      return num
  
  def work(n,k):
      res=10**11
      _n=str(n)
      wid=len(_n)
      vis=dict()
      tmp=0
      cnt=0
      for i in range(wid):
          num=int(_n[i])
          vis.setdefault(num,0)
          if vis[num]==1:
              tmp=tmp*10+num
              continue
          if cnt==k:
              res=check(n,k,tmp,i,_n,wid)
              if res==-1:
                  return work((tmp+1)*(10**(wid-i)),k)
              return res
          cnt+=1
          vis[num]=1
          tmp=tmp*10+num
      return tmp
  
  _t=int(input())
  for _c in range(_t):
      n,k=map(int,input().split())
      print(work(n,k))