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.Scanner;
  
  /**
   *
   * @author eslam
   */
  public class Banknotes {
  
      /**
       * @param args the command line arguments
       */
      public static void main(String[] args) {
          Scanner input = new Scanner(System.in);
          int t = input.nextInt();
          for (int i = 0; i < t; i++) {
              int n = input.nextInt();
              long k = input.nextInt();
              long ans = 0;
              long num = 0;
              long a[] = new long[n];
              for (int j = 0; j < n; j++) {
                  int x = input.nextInt();
                  a[j] = (long) Math.pow(10, x);
              }
              for (int j = 0; j < n-1; j++) {
                  long x = (a[j+1]-a[j])/a[j];
                  long d = (k+1)-num;
                  ans+=Math.min(x, d)*a[j];
                  num+=Math.min(d, x);
              }
              ans+=((k+1)-num)*a[n-1];
              System.out.println(ans);
          }
      }
      
  }