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: for fghn in range(int(input())):
      # n=int(input())
      n,k=list(map(int,input().split()))
      a=list(map(int,input().split()))
      note=0
      value=0
      for i in range(1,n):
          # note+a[i]/a[i-1]-1
          # value+=(a[i]/a[i-1]-1)*(10**a[i])
          if(note+(10**a[i])/(10**a[i-1])-1>k):
              print(int(value)+int((k-note+1)*10**a[i-1]))
              break
          note+=(10**a[i])/(10**a[i-1])-1
          value+=((10**a[i])/(10**a[i-1])-1)*(10**a[i-1])
      else:
          # print("here",)
          print(int(value)+int(k-note+1)*(10**a[-1]))