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: for _ in range(int(input())):
      n, k = map(int, input().split())
      s = str(n)
      if k == 1:
          if int(s[0] * len(s)) >= n:
              print(s[0] * len(s))
          else:
              print(str(int(s[0]) + 1) * len(s))
          continue
  
      dg, prob = set(), False
      for i, v in enumerate(s):
          if v not in dg:
              if len(dg) < k:
                  dg.add(v)
              else:
                  prob = i
                  break
  
      if not prob:
          print(n)
      else:
          res, g = s[prob:], max(dg)
          if int(str(g) * len(res)) >= int(res):
              ans = s[:prob]
              for i in range(prob, len(s)):
                  if s[i] in dg:
                      ans += s[i]
                  else:
                      ans += min(d for d in dg if d > s[i])
                      omp = len(s) - len(ans)
                      ans += min(dg) * omp
                      break
              print(ans)
          else:
              for i in range(prob-1, -1, -1):
                  if s[i] < g:
                      prob = i
                      break
              if s[prob] in s[:prob]:
                  ans = s[:prob]
                  ans += min(d for d in dg if d > s[prob])
                  omp = len(s) - len(ans)
                  ans += min(dg) * omp
  
              else:
                  dg.remove(s[prob])
                  ans = s[:prob]
                  nou = str(int(s[prob]) + 1)
                  ans += nou
  
                  if nou in dg:
                      dg.add('0')
                  omp = len(s) - len(ans)
                  ans += min(dg) * omp
  
              dg, k = set(), k - 1
              for i, v in enumerate(s):
                  if v not in dg:
                      if len(dg) < k:
                          dg.add(v)
                      else:
                          prob = i
                          break
  
              ans2 = s[:prob]
              nou = str(int(s[prob]) + 1)
              ans2 += nou
  
              dg.add(nou)
              omp = len(s) - len(ans2)
              ans2 += min(dg) * omp
              print(min(ans, ans2))
  