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
  import io, os
  input = sys.stdin.buffer.readline
  #input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
  
  t = int(input())
  for _ in range(t):
      n,k = map(int, input().split())
      A = list(map(int, input().split()))
      B = [10**a for a in A]
      cur_max = B[0]*k
      for i in range(1, n):
          a = A[i]
          b = B[i]
          y = B[i]-1
          cnt = 0
          for j in reversed(range(i)):
              q, r = divmod(y, B[j])
              cnt += q
              y = r
          if cnt > k:
              ans = cur_max+1
              break
          else:
              cur_max = (k-cnt+1)*B[i]+B[i]-2
      else:
          print(cur_max+1)
          continue
      print(ans)
  