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
  from collections import Counter
  
  for t in range(int(input())):
      # a, b, c = map(int, input().split())
      # A = list(map(int, input().split()))
      n, k = map(int, input().split())
      L = len(str(n))
      if k == 1:
          for x in range(1, 10):
              cur = int(str(x) * L)
              if cur >= n:
                  print(cur)
                  break
      else:
          for x in range(1, 10):
              cur = int(str(x) * L)
              if cur >= n:
                  break
          # if len(set(n)) == 1:
          #     print(cur)
          #     continue
          #
  
  
          for x in range(10):
              for y in range(x + 1, 10):
                  x, y = str(x), str(y)
                  flag = 0
                  tmp = ""
                  flag2 = 0
                  last = {}
                  for i, c in enumerate(str(n)):
                      if not flag:
                          if c == x or c == y:
                              tmp += c
                          elif x > c:
                              tmp += x
                              flag = 1
                          elif y > c:
                              tmp += y
                              flag = 1
                          else:
                              if x not in last:
                                  flag2 = 1
                              else:
                                  tmp = str(n)[: last[x]] + y + x * (L - 1 - last[x])
                              break
                      else:
                          tmp += x
                      last[c] = i
                  if flag2: continue
                  if tmp[0] == '0': continue
                  cur = min(cur, int(tmp))
          print(cur)
  
  
  