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: from sys import stdin
  import math
  input = stdin.readline
  
  
  # for _ in range(int(input())):
  # 	n, k = map(int, input().split())
  # 	denom = list(map(int, input().split()))
  # 	if(denom[0] != 0):
  # 		print(1)
  # 	elif(k < 9 or len(denom) == 1):
  # 		print(k + 1)
  # 	else:
  # 		req, values, done = 0, [], False
  # 		for i in range(n - 1):
  # 			curr = pow(10, denom[i + 1] - denom[i])
  # 			if(req + curr >= k):
  # 				values.append(curr)
  # 				continue
  # 			else:
  # 				val = pow(10, denom[i]) * k - req + 1
  # 				for j, v in enumerate(values[-1]):
  # 					val += pow(10, denom[j]) * v
  # 				if(len(values) > 0):
  # 					val += pow(10, denom[len(values) -1]) * (values[-1] - 1)
  # 				done = True
  # 				break
  				
  # 		if(not done):
  
  
  
  
  # 			req += curr
  
  
  for _ in range(int(input())):
  	n, k = map(int, input().split())
  	d = list(map(int, input().split()))
  	kk = k+1
  	ans = 0
  	for i in range(len(d)-1):
  		s = pow(10,d[i])
  		c = pow(10,d[i+1]-d[i]) - 1
  		ans += min(c,kk) * s
  		kk -= min(c,kk)
  	ans += kk * pow(10,d[-1])
  	print(ans)
  
  
  
  