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.io.*;
  import java.util.*;
  
  public class AMain {
  	static long myPow(long x, int pw)
  	{
  		long res = 1;
  		for(int i=0;i<pw;i++)
  			res=res*x;
  		return res;
  	}
  	
  	public static void main(String[] args) {
  		try(
  			Scanner in = new Scanner(new BufferedInputStream(System.in));
  			PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));)
  		{
  			int t = in.nextInt();
  			while(t-->0) {
  				int n,k;
  				n = in.nextInt();
  				k = in.nextInt();
  				
  				int[] a = new int[n];
  				for(int i=0;i<n;i++)
  					a[i] = in.nextInt();
  				
  				long res = 0;
  				int left = k+1;
  				for(int i=0;i<n;i++)
  				{
  					long cnt = (i+1<n) ? Math.min(left,myPow(10L, a[i+1]-a[i])-1): (long)left; 
  					res += myPow(10L, a[i])*cnt;
  					left-=cnt;
  				}
  				
  				out.println(res);
  			}
  		}
  		
  	}
  }
  