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 collections import defaultdict, Counter,deque
  from math import sqrt, log10, log, floor, factorial
  from bisect import bisect_left, bisect_right
  from itertools import permutations,combinations
  import sys, io, os
  input = sys.stdin.readline
  # input=io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
  # sys.setrecursionlimit(10000)
  inf = float('inf')
  mod = 10 ** 9 + 7
  def yn(a): print("YES" if a else "NO")
  ceil = lambda a, b: (a + b - 1) // b
  uni=[]
  strings={str(i) for i in range(1,10)}
  for i in strings:
      for j in range(1,11):
          uni.append(int(i*j))
  uni.sort()
  doubleset=set()
  def init(a,b,string):
      if len(string)>=10:
          return
      if string:
          doubleset.add(int(string))
      init(a,b,string+a)
      init(a,b,string+b)
  for i in range(10):
      for j in range(i+1,10):
          init(str(i),str(j),"")
  
  doub=sorted(doubleset)
  
  t=int(input())
  for i in range(t):
      n, k = [int(i) for i in input().split()]
      ncopy = n
      n = str(n)
      s = set(n)
      if len(s) <= k:
          print(n)
          continue
      if k == 1:
          print(uni[bisect_left(uni, ncopy)])
      else:
          print(doub[bisect_left(doub,ncopy)])
  
  
  
  
  
  
  
  