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
  from functools import lru_cache
  
  
  for t in range(int(input())):
      n, k = map(int, input().split())
      s = str(n)
      m = len(s)
      p = [0] * 12
      p[0] = 1
      for i in range(1, 12): p[i] = 10 * p[i - 1]
  
      @lru_cache(None)
      def dp(i, tight, mask):
          if i == m: return 0
          ans = float("inf")
          for j in range(10):
              mask2 = mask | 1 << j
              if bin(mask2).count('1') > k: continue
              if tight and j < int(s[i]): continue
              tight2 = tight and j == int(s[i])
              cur = p[m - 1 - i] * j + dp(i + 1, tight2, mask2)
              if cur < ans:
                  ans = cur
                  break
          return ans
  
      ans = dp(0, 1, 0)
      print(ans)