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: # Educational Codeforces Round 116 (Rated for Div. 2)
  # C. Banknotes
  
  number_tests = int(input(""))
  
  for test in range(number_tests):
      list_input = input("").split()
      number_types = int(list_input[0])
      amount = int(list_input[1]) + 1
      list_types = input("").split()
  
      result = 0
      count = 0
  
      for types in range(number_types):
          if count != 0:
              actual = 10**int(list_types[count])
              previous = 10**int(list_types[count-1])
              diff = int(actual / previous) - 1
              
              if diff >= amount and amount > 0:
                  result += amount * previous
                  amount = 0
              elif amount > diff and amount > 0 and types == number_types - 1: 
                  result += diff * previous
                  amount -= diff
                  result += amount * actual
              elif amount > diff and amount > 0: 
                  result += diff * previous
                  amount -= diff
              
          count += 1
      
      if count == 1:
          print(amount)
      else:
          print(result)