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 sys
  input = sys.stdin.readline
  
  for _ in range (int(input())):
      n,k = [int(i) for i in input().split()]
      sn = [int(i) for i in str(n)]
      s = set()
      ans = []
      for i in sn:
          if i in s:
              ans.append(i)
          elif len(s)<k:
              s.add(i)
              ans.append(i)
          else:
              flag = 0
              for j in range (1,10):
                  if i+j in s:
                      ans.append(i+j)
                      flag = 1
                      break
              if flag:
                  ans.extend([min(ans)]*(len(sn)-len(ans)))
              else:
                  while(len(ans)):
                      if ans.count(ans[-1])==1:
                          s.remove(ans[-1])
                          s.add(ans[-1]+1)
                          ans.append(ans.pop()+1)
                          break
                      else:
                          flag1 = 0
                          for j in range (1,10):
                              if ans[-1]+j in s:
                                  flag1 = ans[-1]+j
                                  break
                          if flag1:
                              ans.pop()
                              ans.append(flag1)
                              break
                          else:
                              ans.pop()
                  if len(s)==k:
                      ans.extend([min(ans)]*(len(sn)-len(ans)))
                  else:
                      ans.extend([0]*(len(sn)-len(ans)))
              break
      for i in ans:
          print(i,end="")
      print()
  