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
  
  inp = sys.stdin.readline
  
  
  def solve():
      n, k = map(int, inp().split())
      a = list(map(int, inp().split()))
      c = [0] * n
      for i in range(n - 1):
          c[i] = 10**(a[i + 1] - a[i]) - 1
      c[n - 1] = 2000000000
      s = 0
      r = []
      #print(c)
      for i in range(n):
          if s + c[i] > k:
              for j in range(i):
                  v = str(c[j])
                  r.append(v + '0'*(len(v) - a[j + 1] - a[j]))
              r.append(str(k + 1 - s))
              break
          s += c[i]
      r.reverse()
      print(''.join(r))
  
  
  def main():
      for i in range(int(inp())):
          solve()
  
  
  if __name__ == '__main__':
      main()
  