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.*;
  
  public class SolnC {
  
  	public static void main(String[] args) {
  		Scanner sc = new Scanner(System.in);
  		int t = sc.nextInt();
  		for(int i=0;i<t;i++) {
  			int n=sc.nextInt();
  			int k=sc.nextInt();
  			k++;
  			int[] a = new int[n];
  			for(int j=0;j<n;j++) {
  				a[j]=sc.nextInt();
  			}
  			long res = 0L;
  			for(int j=0;j<n-1;j++) {
  				if(k>0) {
  					long ans = (long) (Math.pow(10, a[j+1]-a[j])-1);
  					long best = Math.min(ans, k);
  					res=(long) Math.pow(10, a[j])*best+res;
  					k-=best;
  				}
  			}
  			if(k>0) {
  				res=res+(long) Math.pow(10, a[n-1])*k;
  			}
  			System.out.println(res);
  		}
  	}
  
  }
  