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: t=int(input())
  for _ in range(t):
    n,k=map(int,input().split())
    arr=list(map(int,input().split()))
    if n==1:
      print(k+1)
      continue
    prev=0
    cur=0
    k+=1
    for i in range(1,n):
      # diff=arr[i]-arr[i-1]
      if 10**arr[i]//(10**arr[i-1])-1>=k:
        cur=cur+(10**arr[i-1])*(k)
        k=0
        break
      else:
        prev=(10**arr[i])//(10**arr[i-1])-1
        k-=prev
        
        cur=cur+(10**arr[i-1])*(prev)
        
    if k:
      
      cur=cur+10**arr[-1]*(k)
    print(cur)