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: from itertools import product
  from bisect import bisect_left
  from sys import stdin
  input = stdin.readline
  
  l1 = [0]
  s = set()
  l2 = []
  def kone():
      for i in range(1,10):
          for j in range(1,11):
              l1.append(int(str(i)*j))
      l1.sort()
  
  def ktwo():
      s = set(l1)
      for i in range(9):
          for j in range(i+1,10):
              for k in range(2,11):
                  for v in product(str(i)+str(j),repeat=k):
                      val = "".join(v)
                      if val not in s:
                          l2.append(int(val))
      l2.sort()
  
  kone()
  ktwo()
  for _ in range(int(input())):
      n,k = map(int,input().split())
      if k==1:
          idx = bisect_left(l1,n)
          print(l1[idx])
      else:
          idx = bisect_left(l2,n)
          print(l2[idx])