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.*;
  import java.io.*;
  import java.math.BigInteger;
  
  
  public class Solution {
  	
  	
  	
  	 public static void main(String[] args) {
  		 Scanner in = new Scanner(System.in);
  		 int n = in.nextInt();
  		 for(int i = 0; i<n; i++) {
  			 int cases = in.nextInt();
  			 int t = in.nextInt();
  			 int[] berlins = new int[cases];
  			 for(int j = 0; j<cases; j++) {
  				 berlins[j]=in.nextInt();
  				 //System.out.print(berlins[j] + " ");
  			 }
  			// System.out.println();
  			 int[] diffArray = new int[cases-1];
  			 for(int j = 0; j<cases-1; j++) {
  				 diffArray[j] = (int)Math.pow(10, berlins[j+1]-berlins[j])-1;
  				 //System.out.print(diffArray[j] + " ");
  			 }
  			 //System.out.println();
  			 int copy = 0;
  			 for(int j = 0; j<cases-1; j++) {
  				 t-=diffArray[j];
  				 if(t<0) {
  					 copy = j;
  					 t+=diffArray[copy];
  					 j=cases;
  				 }
  				 else copy=j+1;
  			 }
  			 System.out.print(t+1);
  			 for(int j = 0; j<berlins[copy]; j++) {
  				 System.out.print(9);
  			 }
  			 System.out.println();
  			 /*String test = String.format("%.0f",((t+2)*(Math.pow(10, berlins[copy]))-1));
  			 BigInteger ans = new BigInteger(test);
  			 ans.subtract(new BigInteger("1"));
  			 System.out.println(ans);*/
  		 }
  		 
  	}
  }