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: t = int(input())
  
  while t:
      num, k = [int(tok) for tok in input().split()]
      ans = 111111111111111
      num = str(num)
      n = len(num)
      s=set()
      for ch in num:
          s.add(ch)
      
      if (len(s) <= k):
          print(int(num))
      else:
  
          for ind in range(0, n):
              if(num[ind] == '9'):
                  continue
              
              done = set()
              for i in range(0, ind):
                  done.add(num[i])
                  
              if (len(done) > k):
                  continue
              
              elif (len(done) == k):
                  to_fill = None
                  mi = '9'
                  for el in done:
                      mi = min(mi, el)
                      if(el > num[ind]):
                          if to_fill is None:
                              to_fill = el
                          else:
                              to_fill = min(to_fill, el)
                              
                  if(to_fill is not None):
                      ans = min(ans, int(num[:ind] + to_fill + mi*(n-ind-1)))
                          
              else:
                  mi = '9'
                  for i in range(0, 9):
                      if(str(i) > num[ind]):
                          mi = str(i)
                          break
                      
                  done.add(mi)
                  if(len(done) == k):
              
                      ans = min(ans, int(num[:ind] + mi + min(done)*(n-ind-1)))
                  else:
                      ans = min(ans, int(num[:ind] + mi + '0'*(n-ind-1)))
                      
              # print(ind, ans)    
                  
          print(ans)
      
      
      t-=1