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 sys import stdin
  input=stdin.readline
  rn=lambda:int(input())
  rns=lambda:map(int,input().split())
  rl=lambda:list(map(int,input().split()))
  rs=lambda:input().strip()
  YN=lambda x:print('YES') if x else print('NO')
  mod=10**9+7
  
  for _ in range(rn()):
      n,k=rns()
      a=rl()
      b=[]
      for i in range(n-1):
          b.append(10**(a[i+1]-a[i])-1)
      ans=0
      while len(b)>0 and k>=b[0]:
          ans+=b[0]*10**a[0]
          k-=b.pop(0)
          a.pop(0)
  
      if k>=0:
          if len(b)==0:
              last=a.pop(0)
              ans+=(k+1)*10**last
          else:
              ans+=(k+1)*10**a.pop(0)
      print(ans)