description: In Berland, n different types of banknotes are used. Banknotes of the i-th type have denomination 10^{a_i} burles (burles are the currency used in Berland); the denomination of banknotes of the first type is exactly 1.
  
  Let's denote f(s) as the minimum number of banknotes required to represent exactly s burles. For example, if the denominations of banknotes used in Berland are 1, 10 and 100, then f(59) = 14: 9 banknotes with denomination of 1 burle and 5 banknotes with denomination of 10 burles can be used to represent exactly 9 ⋅ 1 + 5 ⋅ 10 = 59 burles, and there's no way to do it with fewer banknotes.
  
  For a given integer k, find the minimum positive number of burles s that cannot be represented with k or fewer banknotes (that is, f(s) > k).
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 10^4) — number of test cases.
  
  The first line of each test case contains two integers n and k (1 ≤ n ≤ 10; 1 ≤ k ≤ 10^9).
  
  The next line contains n integers a_1, a_2, ..., a_n (0 = a_1 < a_2 < ... < a_n ≤ 9).
  
  Output
  
  For each test case, print one integer — the minimum positive number of burles s that cannot be represented with k or fewer banknotes.
  
  Example
  
  Input
  
  
  4
  3 13
  0 1 2
  2 777
  0 4
  3 255
  0 1 3
  10 1000000000
  0 1 2 3 4 5 6 7 8 9
  
  
  Output
  
  
  59
  778
  148999
  999999920999999999
solution: import io
  import os
  
  FILE_INPUT_MODE = 0  # 0 for BytesIO, 1 for input.txt, -1 for stdin, input()
  if FILE_INPUT_MODE >= 0:
  
      if FILE_INPUT_MODE == 1:
          CURRENT_DIR = os.path.dirname(os.path.abspath(__file__))
          FD = os.open(f"{CURRENT_DIR}\input.txt", os.O_RDONLY)
      elif FILE_INPUT_MODE == 0:
          FD = 0
      DATA = io.BytesIO(os.read(FD, os.fstat(FD).st_size))
      if FD != 0:
          os.close(FD)
  
      def input():
          return DATA.readline().decode().rstrip('\r\n')
  
  
  def solve(n, k, a, d):
      ans = 0
      cnt = 0
      for i in range(1, n):
          x = d[a[i] - a[i - 1]] - 1
          if k + 1 - cnt <= x:
              ans += (k + 1 - cnt) * d[a[i - 1]]
              return ans
          ans += x * d[a[i - 1]]
          cnt += x
      ans += (k + 1 - cnt) * d[a[n - 1]]
      return ans
  
  
  def main():
      t = int(input())
      d = [1]
      for i in range(10):
          d.append(d[-1] * 10)
      for _ in range(t):
          n, k = [int(word) for word in input().strip().split()]
          a = [int(word) for word in input().strip().split()]
          print(solve(n, k, a, d))
  
  
  if __name__ == "__main__":
      main()
  