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: #!/usr/bin/env python3
  # from typing import *
  
  import sys
  import io
  import math
  import collections
  import decimal
  import itertools
  import bisect
  import heapq
  
  
  def input():
      return sys.stdin.readline()[:-1]
  
  
  # sys.setrecursionlimit(1000000)
  
  # _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
  
  # """
  # sys.stdin = io.StringIO(_INPUT)
  
  INF = 10**10
  
  def solve(N, K, A):
      ans = 0
      A += [20]
      K += 1
      for i in range(N):
          if K < 10**(A[i+1]-A[i]):
              ans += K*(10**A[i])
              return ans
          else:
              v = (10**(A[i+1]-A[i])-1)
              ans += v*(10**A[i])
              K -= v
  
      return ans
  
  T0 = int(input())
  for _ in range(T0):
      N, K = map(int, input().split())    
      A = list(map(int, input().split()))
      print(solve(N, K, A))