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: t=int(input())
  k1=[0]
  for i in range(1, 10):
      for j in range(1, 11):
          e=str(i)*j
          e=int(e)
          k1.append(e)
  k1=sorted(k1)
  
  k2=[]
   
  def ad(i, j, s):
      if len(s)==10:
          if s!=s[0]*len(s):
              k2.append(int(s))
      else:
          if s!='' and s!=s[0]*len(s):
              k2.append(int(s))
          if not (i=='0' and s==''):
              ad(i, j, s+i)
          ad(i, j, s+j)
   
  for i in range(10):
      for j in range(i+1, 10):
          ad(str(i), str(j), '')
  
  
  k2+=k1
  k2=sorted(k2)
   
  def binsearch(l, r, arr, n):
      while l+1<r:
          mid=(l+r)//2
          if arr[mid]<=n:
              l=mid
          else:
              r=mid
  
      if n==arr[l]:
          return l
      return r
   
  for i in range(t):
      a,b=map(int, input().split())
      if b==1:
          ans=binsearch(-1, len(k1), k1, a)
          print(k1[ans])
      else:
          ans=binsearch(-1, len(k2), k2, a)
          print(k2[ans])