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())
  
  for t in range(T):
      n, k = map(int, input().split())
      s = str(n)
      if k == 1:
          t = s[0] * len(s)
          if t >= s:
              print(t)
          else:
              t = chr(ord(s[0]) + 1) * len(s)
              print(t)
      else:
          cur, cnt = s[0], 0
          ans = ''
          ans += s[0]
          
          isUsed = [False for _ in range(10)]
          isUsed[ord(s[0]) - ord('0')] = True
          mx = ord(s[0]) - ord('0')
          for i in range(1, len(s)):
              x = ord(s[i]) - ord('0')
              if s[i] == s[i - 1]:
                  ans += s[i]
              else:
                  if not isUsed[x]:
                      k -= 1
                  if k == 1:
                      break 
                  else:
                      ans += s[i]
                      isUsed[x] = True
                      mx = max(mx, x)
  
          cnt = len(ans)
          for i in range(len(s) - cnt):
              for d in range(10):
                  t = ans + chr(d + ord('0')) + chr(max(d, mx) + ord('0')) * (len(s) - cnt - i - 1)
                  if not isUsed[d] and k == 1 and t >= s:
                      ans += chr(d + ord('0'))
                      isUsed[d] = True
                      mx = max(mx, d)
                      k -= 1
                      break
                  elif isUsed[d] and t >= s:
                      ans += chr(d + ord('0'))
                      break   
          print(ans)     