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;
  
  public class Main {
      public static Scanner sc = new Scanner(System.in);
      public static void main(String[] args) {
          int t = sc.nextInt();
          while (t > 0) {
              run();
              t --;
          }
      }
  
      public static void run() {
          int n = sc.nextInt();
          int k = sc.nextInt();
          int[] a = new int[10];
          int[] r = new int[10];
          int[] p = new int[10];
  
          p[0] = 1;
          for(int i = 1; i < 10; i ++) {
              p[i] = p[i-1] * 10;
          }
          for(int i = 0; i < n; i ++) {
              a[i] = sc.nextInt();
          }
          k ++;
          for(int i = 1; i <= n; i ++) {
              if(i == n) {
                  if(k > 0) {
                      r[a[n-1]] += k;
                  }
                  break;
              }
              int x = Math.min(k, p[a[i]-a[i-1]]-1);
              r[a[i-1]] = x;
              k -= x;
          }
          long res = 0L;
          for(int i = 0; i < 10; i ++) {
              res += 1l * r[i] * p[i];
  //            System.out.println(r[i] + " " + p[i]);
          }
          System.out.println(res);
      }
  }