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
  import io, os
  #input = sys.stdin.buffer.readline
  input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
  
  t = int(input())
  
  X1 = []
  for i in range(1, 10):
      for j in range(1, 11):
          x = str(i)*j
          x = int(x)
          X1.append(x)
  
  import itertools
  X2 = []
  for i in range(10):
      for j in range(10):
          if i == j:
              continue
          A = [str(i), str(j)]
          for l in range(1, 11):
              for p in itertools.product(A, repeat=l):
                  if p[0] == '0':
                      continue
                  p = list(p)
                  p = ''.join(p)
                  p = int(p)
                  X2.append(p)
  X1.sort()
  X2.sort()
  import bisect
  for _ in range(t):
      n, k = map(int, input().split())
      if k == 1:
          j = bisect.bisect_left(X1, n)
          print(X1[j])
      else:
          j = bisect.bisect_left(X2, n)
          print(X2[j])
  