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: #for p in [(x, 5) for x in range(344660200, 344660300)]:
  for _ in range(int(input())):
      #n, k = p[0], p[1]
      #n = str(n)
      n, k = input().split()
      k = int(k)
      a = []
      ans = 0
      it = False
      for i in range(len(n)):
          if it:
              if k:
                  ans = ans * 10
              else:
                  ans = ans * 10 + min(a)
              continue
          #if n[i] == max(n[i:]) or k > 1 or k > 0 and n[i] in a or a and max(a) >= int(max(n[i:])):
          j = i
          while j < len(n):
              if n[j] > n[i]:
                  j += 1
                  break
              j += 1
          j -= 1
          kk = j
          while kk < len(n):
              if n[kk] < n[j]:
                  kk += 1
                  break
              kk += 1
          kk -= 1
          if n[i] == max(n[i:]) or k > 1 or (k > 0 and (int(n[i]) in a)) or a and max(a) >= int(max(n[j:kk + 1])) or i == len(n) - 1 or a and min([int(n[x]) for x in range(i + 1, j + 1)]) < max(a) or min([int(n[x]) for x in range(i + 1, j + 1)]) < int(n[i]):
              d = int(n[i])
          else:
              d = int(n[i]) + 1
          if d not in a and k:
              k -= 1
              a.append(d)
              a.sort()
              ans = ans * 10 + d
          else:
              for x in a:
                  if x >= d:
                      ans = ans * 10 + x
                      break
          if ans % 10 > int(n[i]):
              it = True
      print(ans)
  