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.io.PrintWriter;
  import java.util.ArrayList;
  import java.util.Arrays;
  import java.util.PriorityQueue;
  import java.util.StringTokenizer;
  
  public class Main {
  
      public static void main(String[] args) {
          FastReader fastReader = new FastReader();
          PrintWriter printWriter = new PrintWriter(System.out);
  
          int t = fastReader.nextInt();
          while (t-- > 0) {
              int n = fastReader.nextInt();
              long k = fastReader.nextLong() + 1;
  
              int[] arr = new int[n];
              for (int i = 0; i < n; i++) {
                  arr[i] = fastReader.nextInt();
              }
  
              long ans = 0;
  
              int i = 0;
              while (k > 0 && i+1 < n) {
                  long diff = Math.min((long)Math.pow(10L, arr[i+1] - arr[i]) - 1, k);
  
                  ans += (long)Math.pow(10L, arr[i]) * diff;
                  k -= diff;
  
                  i++;
              }
  
              ans += (long)Math.pow(10L, arr[n-1]) * k;
  
              printWriter.println(ans);
          }
  
          printWriter.close();
      }
  
      static class FastReader
      {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader()
          {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
  
          String next()
          {
              while (st == null || !st.hasMoreElements())
              {
                  try
                  {
                      st = new StringTokenizer(br.readLine());
                  }
                  catch (IOException e)
                  {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt()
          {
              return Integer.parseInt(next());
          }
  
          long nextLong()
          {
              return Long.parseLong(next());
          }
  
          double nextDouble()
          {
              return Double.parseDouble(next());
          }
  
          String nextLine()
          {
              String str = "";
              try
              {
                  str = br.readLine();
              }
              catch (IOException e)
              {
                  e.printStackTrace();
              }
              return str;
          }
      }
  }
  