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: from math import sqrt,gcd,ceil,floor,log,factorial
  from itertools import permutations,combinations
  from collections import Counter, defaultdict
  import collections,sys,threading
  import collections,sys,threading
  from bisect import *
  from heapq import *
  #sys.setrecursionlimit(10**9)
  #threading.stack_size(10**8)
  #input=sys.stdin.readline
  
  def ii(): return int(input())
  def si(): return input()
  def mi(): return map(int,input().split())
  def msi(): return map(str,input().split())
  def li(): return list(mi())
  
  t=ii()
  for _ in range(t):
      n,k=mi()
      a=li()
      av=[0]*(n-1)
      pre=[0]*(n-1)
      money=[0]*(n-1)
      
      
              
      for i in range(n-1):
          av[i] = (pow(10,a[i+1])-pow(10,a[i]))
          pre[i]=av[i]//pow(10,a[i]) + pre[i-1]
          money[i]=money[i-1]+av[i]
          
      
      flag=0   
      #print(*pre)
      #print(*money)
      for i in range(n-2,-1,-1):
          if k>=pre[i]:
              req=k-pre[i]+1
              flag=1
              print(req*(pow(10,a[i+1])) + (money[i]))
              break
          
      if flag==0:
          print(k+1)
          
              
          
      
              
              
      
      