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 sys
  import os.path
  from collections import *
  import math
  import bisect
  import heapq as hq
  from fractions import Fraction
  from random import randint
  
  if (os.path.exists('input.txt')):
      sys.stdin = open("input.txt", "r")
      sys.stdout = open("output.txt", "w")
  
  input = sys.stdin.readline
  
  ##########################################################
  
  def solve():
      n, k = [int(x) for x in input().split()]
      arr = [int(x) for x in input().split()]
      n += 1
      k += 1
      arr.append(20)
      brr = [0] * n
      for i in range(n):
          brr[i] = 10 ** arr[i]
      res = 0
      for i in range(n - 1):
          if brr[i] * k < brr[i + 1]:
              res += brr[i] * k
              break
          k -= (brr[i + 1] - brr[i]) // (10 ** arr[i])
          # if(i == 0):
          #     res += brr[i + 1] - 2 * brr[i]
          #     k += 1
          # else:
          res += brr[i + 1] - brr[i]
          # print(res, k)
  
      print(res)
  
  
  t = int(input())
  while t:
      t -= 1
      solve()
  
  ##########################################################
  