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 main(t):
   
   
      n,k = map(int, input().split())
      s = str(n)
      m = len(s)
   
   
      dic = {}
   
      ans = ""
   
      for i in range(m):
   
          if s[i] in dic:
              dic[s[i]] += 1
          else:
              dic[s[i]] = 1
        
   
   
   
          if len(dic)==k and i<m-1:
              if s[i+1:] >  max(dic)*(m-i-1):
                  dic[s[i]] -= 1
                  if dic[s[i]]==0:  del dic[s[i]] 
                  if str(int(s[i])+1) in dic and len(dic)<k:
                      ans += str(int(s[i])+1)
                      ans += '0'*(m-i-1)
                  else:
                      dic[str(int(s[i])+1)] = 1
                      ans += str(int(s[i])+1)
                      ans += min(dic)*(m-i-1)
                  break
   
              else:
                  ans += s[i]
                  broken = i+1
                  flag = True
                  for j in range(i+1,m):
                      if s[j] < max(dic): broken = j
                      if s[j] not in dic:  
                          flag = False
                          break
                  if flag:  broken=m 
                  ans += s[i+1:broken]
                  if broken==m:  break
   
                  for d in range(int(s[broken])+1,10):
                      if str(d) in dic: break
                  dic[str(d)] = 1
                  ans += str(d)
                  ans += min(dic)*(m-len(ans))
                  break
   
                    
   
          ans += s[i] 
   
      print(ans)
  
  T = int(input())
  t = 1
  while t<=T:
      main(t)
      t += 1 
  