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.Arrays;
  import java.util.Scanner;
  import java.util.stream.Collectors;
  import java.util.stream.Stream;
  
  public class Banknotes {
  
      public static void main(String[] args) {
          Scanner sc = new Scanner(System.in);
          int tests = sc.nextInt();
          sc.nextLine();
          for(int i = 0;i<tests;i++) {
              String firstLine = sc.nextLine();
              String[] banknotes = sc.nextLine().split(" ");
              int k = Integer.parseInt(firstLine.split(" ")[1]);
              int[] banks = Arrays.stream(banknotes).mapToInt(Integer::parseInt).toArray();
              solution(k, banks);
          }
      }
  
      public static void solution(int k, int[] banknotes) {
          k++;
          int[] poweredNumbs = Arrays.stream(banknotes).map(num -> (int) Math.pow(10, num)).toArray();
          long result = 0;
          int remaining = k;
          for(int i = 0;i<banknotes.length;i++) {
              int maxPotential = i + 1 < banknotes.length ? poweredNumbs[i + 1] / poweredNumbs[i] - 1 : remaining;
              if(maxPotential > remaining || banknotes.length - 1 == i) {
                  result += (long) poweredNumbs[i] * remaining;
                  System.out.println(result);
                  return;
              } else {
                  result += poweredNumbs[i] * maxPotential;
                  remaining -= maxPotential;
              }
          }
  
      }
  }
  