description: It is a simplified version of problem F2. 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 ≤ 2).
  
  Output
  
  For each test case output on a separate line x — the minimum k-beautiful integer such that x ≥ n.
  
  Example
  
  Input
  
  
  4
  1 1
  221 2
  177890 2
  998244353 1
  
  
  Output
  
  
  1
  221
  181111
  999999999
solution: import sys
  input = sys.stdin.readline
  
  bits = [[] for _ in range(11)]
  for bit in range(1 << 10):
      se = set()
      for i in range(10):
          if bit >> i & 1:
              se.add(i)
      bits[len(se)].append(se)
  
  def main():
      n, k = input().strip().split()
      k = int(k)
      l = len(n)
      times = (10 ** l - 1) // 9
      min_ = 10 ** 20
      for se in bits[k]:
          if max(se) * times < int(n):
              continue
          tmp = -1
          nex = [-1] * 10
          for i in range(9, -1, -1):
              if i in se:
                  tmp = i
              nex[i] = tmp
          
          ans = []
          flg = False
          for i in range(l):
              if flg:
                  ans.append(nex[0])
                  continue
              s = int(n[i])
              t = nex[s]
              if t == s:
                  ans.append(s)
              elif t != -1:
                  ans.append(t)
                  flg = True
              else:
                  cnt = 1
                  while 1:
                      s = ans[-1]
                      ans.pop()
                      if nex[s + 1] != -1:
                          ans.append(nex[s + 1])
                          break
                      cnt += 1
                  for _ in range(cnt):
                      ans.append(nex[0])
                  flg = True
          min_ = min(min_, int("".join(map(str, ans))))
      print(min_)
      
  for _ in range(int(input())):
      main()