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 *
  #from bisect import *
  #from collections import *
  #from random import *
  #from decimal import *"""
  #from heapq import *
  #from itertools import *
  import sys
  input=sys.stdin.readline
  def inp():
      return int(input())
  def st():
      return input().rstrip('\n')
  def lis():
      return list(map(int,input().split()))
  def ma():
      return map(int,input().split())
  t=inp()
  while(t):
      t-=1
      n,k=ma()
      a=lis()
      if(n==1):
          print(k+1)
          continue
      fl=0
      k+=1
      res=''
      for i in range(n-1):
          ra=(10**(a[i+1]))-1
          curmo=0
          num=ra
          for j in range(i,-1,-1):
              ra1=ra//(10**a[j])
              curmo+=ra1
              ra=ra%(10**a[j])
          if(curmo<k):
              continue
          else:
              z1=10**(a[i])-1
              res=str(z1)
              moves=0
              ra=z1
              for j in range(i-1,-1,-1):
                  ra1=ra//(10**a[j])
                  moves+=ra1
                  ra=ra%(10**a[j])
              rem=k-moves
              if(int(res)!=0):
                  res=str(rem)+res
              else:
                  res=str(rem)
              fl=1
              break
      if(fl==0):
              z1=10**(a[-1])-1
              res=str(z1)
              moves=0
              ra=z1
              for j in range(n-2,-1,-1):
                  ra1=ra//(10**a[j])
                  moves+=ra1
                  ra=ra%(10**a[j])
              rem=k-moves
              if(int(res)):
                  res=str(rem)+res
              else:
                  res=str(rem)
              fl=1
      print(int(res))
          
                  
              
          
          
  