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 sys import stdin,stdout
  def INPUT():return list(int(i) for i in stdin.readline().split())
  def inp():return stdin.readline()
  def out(x):return stdout.write(x)
  import math
  INT_MAX=10**13
  from bisect import bisect_left
  import sys
  MOD=10**9+7
  inf=MOD**2
  from decimal import*
  #===================================================================
  # import kivy
  # from kivy.app import App
  # from kivy.uix.label import Label
  #
  # class A(App):
  #     def build(self):
  #         return Label(text="Hello ")
  # if __name__=='__main__':
  #     A().run()
  
  def calc(n):
      A=list(int(ele) for ele in list(str(n)))
      bool=False
  
      for i in range(1,len(A)):
          if A[i]!=A[i-1]:
              if A[i]>A[i-1]:
                  bool=True
                  x=A[i-1]
              break
      if bool is True:
          return str(x+1)*len(A)
      else:
          return str(A[0])*len(A)
  def special(a,b,n):
      l=len(str(n))
      X=[]
      for i in range(1<<l):
          ans=""
          for j in range(l):
              if (1<<j)&i:
                  ans+=str(a)
              else:
                  ans+=str(b)
          X.append(ans)
      return X
  D=[]
  for k in range(1,10):
      first=[]
      for i in range(10):
          for j in range(i+1,10):
              for ele in special(i,j,10**k):
                  if int(ele)>=10**k:
                      first.append(int(ele))
      first.sort()
      D.append(first)
  
  
  
  
  
  
  
  
  
  def calc2(n):
      if n==10**9 or len(set(list(str(n))))==1 or n<=100:
          return n
      l=len(str(n))
      X=D[l-2]
      ans=inf
      return(X[bisect_left(X,n)])
  
  
  for _ in range(int(input())):
      n,k=map(int,input().split())
      if k==1:
          out(str(calc(n))+'\n')
      else:
          out(str(calc2(n))+'\n')
  