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 math import *
  from bisect import *
  #from collections import *
  #from random import *
  #from decimal import *"""
  #from heapq import *
  import sys
  input=sys.stdin.readline
  def inp():
      return int(input())
  def st():
      return input().rstrip('\n')
  def lis():
      return list(map(int,input().split()))
  def ma():
      return map(int,input().split())
  t=inp()
  fi=[int('1'*10)]
  for i in range(1,10):
      for j in range(1,11):
          x=str(i)*j
          fi.append(int(x))
  se=[int('1'*10),10**9]
  for i in range(2,10):
      for j in range(10):
          for k1 in range(10):
              for k2 in range(2**i):
                  nu=''
                  z=bin(k2)[2:]
                  z=z.zfill(i)
                  for k3 in range(i):
                      if(z[k3]=='1'):
                          nu+=str(j)
                      else:
                          nu+=str(k1)
                  se.append(int(nu))
  fi.sort()
  se.sort()
  fk=float('inf')
  while(t):
      t-=1
      n,k=ma()
      if(k==1):
          z=bisect_left(fi,n)
          print(fi[z])
          
      else:
          z=bisect_left(se,n)
          print(se[z])
  