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: a=[]
  for i in range(10):
      for j in range(i,10):
          for m in range(1,11):
              for k in range(2**m):
                  b=bin(k)[2:]
                  b='0'*(m-len(b))+b
                  y=''
                  for l in b:
                      if l=='0':
                         y+=str(i)
                      else:
                         y+=str(j)
                  if int(y)>0:
                      a.append(int(y))
  a.sort()
  for i in range(int(input())):
      n,k=input().split()
      k=int(k)
      if k==1:
          if int(n[0]*len(n))>=int(n):
              print(n[0]*len(n))
          elif n[0]=='9':
              print('1'*(len(n)+1))
          else:
              print(str(int(n[0])+1)*len(n))
      else:
          l=-1
          r=len(a)
          n=int(n)
          while r-l>1:
              m=(l+r)//2
              if a[m]<n:
                  l=m
              else:
                  r=m
          print(a[r])