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,os,io
  import math 
  from collections import defaultdict
  def ii():
      return int(input())
  def li():
      return list(map(int,input().split()))
  if(os.path.exists('input.txt')):
      sys.stdin = open("input.txt","r") ; sys.stdout = open("output.txt","w") 
  else:
      input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
  
  def solve():
      n,k = li()
  
      a = li()
      x = []
      k+=1
      a.sort()
      for i in range(1,n):
          x.append(a[i]-a[i-1])
      ans = 0
      notes = 0
      for i in range(len(x)):
          if k>0:
              ans += (10**a[i]) * min(10**(x[i])-1,k)
              k-=min(10**(x[i])-1,k)
      print(ans+k*(10**a[-1]))
          
      
      
  
  t = 1
  t = ii()
  for _ in range(t):
      solve()
  