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 java.util.Scanner;
  
  public class B750 {
  
  	public static void main(String[] args) {
  		Scanner sc = new Scanner(System.in);
  		int t = sc.nextInt();
  		int[] pp = new int[10];
  		for (int p = 1, i = 0; i < 10; i++) {
  			pp[i] = p;
  			p *= 10;
  		}
  		
  		while (t-- > 0) {
  			int n = sc.nextInt();
  			int k = sc.nextInt();
  			int[] aa = new int[n];
  			
  			for (int i = 0; i < n; i++) aa[i] = sc.nextInt();
  			
  			long ans = 0;
  			for (int i = 0; i < n; i++) {
  				int ki = (int) (i == n - 1 ? k + 1 : Math.min(k+1, pp[aa[i+1] - aa[i]] - 1));
  				ans += (long) ki * pp[aa[i]];
  				if (ki == k + 1) break;
  				k -= ki;
  			}
  			System.out.println(ans);
  		}
  		
  		sc.close();
  	}
  
  }
  