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: # Problem: C. Banknotes
  # Contest: Codeforces - Educational Codeforces Round 116 (Rated for Div. 2)
  # URL: https://codeforces.com/contest/1606/problem/C
  # Memory Limit: 256 MB
  # Time Limit: 2000 ms
  # 
  # KAPOOR'S
  
  from sys import stdin, stdout 
  
  def INI():
  	return int(stdin.readline())
  	
  def INL():
  	return [int(_) for _ in stdin.readline().split()]
  	
  def INS():
  	return stdin.readline()
  
  def MOD():
      return pow(10,9)+7
  	
  def OPS(ans):
  	stdout.write(str(ans)+"\n")
  	
  def OPL(ans):
  	[stdout.write(str(_)+" ") for _ in ans]
  	stdout.write("\n")
  
  
  	
  if __name__=="__main__":
  	for _ in range(INI()):
  		n,k=INL()
  		X=INL()
  		k+=1
  		
  		ans=0
  		_=0
  		while 0<k and _+1<n:
  			ans+=10**X[_]*(min(10**(X[_+1]-X[_])-1,k))
  			k-=min(10**(X[_+1]-X[_])-1,k)
  			_+=1
  
  		OPS(ans+k*(10**X[n-1]))