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.BufferedReader;
  import java.io.IOException;
  import java.io.InputStreamReader;
  import java.util.StringTokenizer;
  
  public class BurlBanknotes {
  	public static void main(String[] args) throws IOException{
  		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
  		StringTokenizer st;
  		int n;
  		int t = Integer.parseInt(br.readLine());
  		int[] k = new int[t];
  		int[][] a = new int[t][];
  		for (int i = 0; i < t; i++) {
  			st = new StringTokenizer(br.readLine());
  			n = Integer.parseInt(st.nextToken());
  			k[i] = Integer.parseInt(st.nextToken());
  			a[i] = new int[n];
  			st = new StringTokenizer(br.readLine());
  			for (int j = 0; j < n; j++) {
  				a[i][j] = Integer.parseInt(st.nextToken());
  			}
  		}
  		
  		long c;
  		int j;
  		for (int i = 0; i < t; i++) {
  			c = 0;
  			if(a[i].length == 1 || k[i] + 1 < Math.pow(10, a[i][1] - a[i][0])) {
  				System.out.println(k[i] + 1);
  				continue;
  			}
  			else {
  				c += (long)Math.pow(10, a[i][1] - a[i][0]) - 2;
  				k[i] -= Math.pow(10, a[i][1] - a[i][0]) - 2;
  			}
  			for (j = 1; j < a[i].length-1; j++) {
  				if(c + k[i] * (long) Math.pow(10, a[i][j]) + 1 >= (long) Math.pow(10, a[i][j+1])) {
  					c += (long) (Math.pow(10, a[i][j+1]-a[i][j]) - 1) * (long) Math.pow(10, a[i][j]);
  					k[i] -= Math.pow(10, a[i][j+1]-a[i][j]) - 1;
  				}
  				else {
  					c += k[i] * (long) Math.pow(10, a[i][j]);
  					k[i] = 0;
  					break;
  				}
  			}
  			if(j == a[i].length-1 && k[i] != 0) {
  				c += k[i] * (long) Math.pow(10, a[i][j]);
  			}
  			System.out.println(c+1);
  		}
  	}
  }
  