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: def recurs(i,n,dist,nw,k,orig):
      if dist > k:
          return ""
      if nw[0:i] < orig[0:i]:
          return ""
      if i>=n:
          return nw
      c = ['0','1','2','3','4','5','6','7','8','9']
      for item in c:
          nw = nw[:i] + item + nw[i+1:]
          f = False
          for j in range(i):
              if nw[j] == item:
                  f = True
                  break
          
          cc = recurs(i+1,n,dist+(f == False),nw,k,orig)
          if cc != "":
              return cc
      return ""
  
  t = int(input())
  for i in range(t):
      n,k = map(int,input().split())
      n = str(n)
      nw = "0"*len(n)
      print(recurs(0,len(n),0,nw,k,n))
  