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.*;
  import java.util.*;
  public class banknotes {
      public static void main (String[] args) throws IOException {
          BufferedReader input = new BufferedReader(new InputStreamReader (System.in));
          PrintWriter out = new PrintWriter(System.out);
          int T = Integer.parseInt(input.readLine());
          StringTokenizer st;
          int n;
          int k;
          for (int t=0;t<T;t++) { // read in test cases
              // read in data
              st = new StringTokenizer(input.readLine());
              n = Integer.parseInt(st.nextToken());
              k = Integer.parseInt(st.nextToken());
              int[] vals = new int[n];
              st = new StringTokenizer(input.readLine());
              for (int i=0;i<n;i++) vals[i] = (int) Math.pow(10, Integer.parseInt(st.nextToken()));
              // solve using greedy algorithm (okay because each bill is an exact multiple of lower bills)
              int goal_num = k+1;
              long sum = 0L;
              int ind = 0;
              while (goal_num > 0) {
                  int num_bills;
                  if (ind < n-1) { num_bills = Math.min(vals[ind+1]/vals[ind] - 1, goal_num); }
                  else num_bills = goal_num;
                  goal_num -= num_bills;
                  sum += (long) vals[ind] * num_bills;
                  if (ind < n-1) ind += 1;
              }
              out.println(sum); }
          out.close();
  
      }
  }
  