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: mod = 1000000007
  
  t = int(input())
  
  p = [1,10,100,1000,10000,100000,1000000,10000000,100000000,1000000000]
  
  while (t):
  
  	n,k = list(map(int,input().split()))
  	a = list(map(int,input().split()))
  	
  	a = [p[x] for x in a]
  	k += 1
  
  	ans = 0
  
  	for i in range(n-1):
  		s = a[i+1]//a[i]-1
  		if (k >= s):
  			ans += s * a[i]
  			k -= s
  		else:
  			ans += k * a[i]	
  			k = 0
  			break
  
  	ans += a[-1] * k
  
  	print(ans)
  
  	t -= 1
  		
  	
  	
  		
  		
  	
  	