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: /*
          "Everything in the universe is balanced. Every disappointment
                  you face in life will be balanced by something good for you!
                          Keep going, never give up."
  
  
  */
  
  
  import java.util.*;
  import java.lang.*;
  import java.io.*;
  
  public class Solution {
      public static void main(String[] args) throws java.lang.Exception {
          out = new PrintWriter(new BufferedOutputStream(System.out));
          sc = new FastReader();
  
          int test = sc.nextInt();
          for (int t = 0; t < test; t++) {
              solve();
          }
          out.close();
      }
  
      private static void solve() {
          int n = sc.nextInt();
          long k = sc.nextLong();
          long[] denomination = new long[n];
          for (int i = 0; i < n; i++) {
              long p = sc.nextLong();
              denomination[i] = (long) Math.pow(10, p);
          }
          long banknotesUsed = 0, minBurles = 0;
          for (int i = 0; i < n; i++) {
              long canUse = (i == n - 1) ? Long.MAX_VALUE / 3 : (denomination[i + 1] - denomination[i]) / denomination[i];
              if (banknotesUsed + canUse > k) {
                  minBurles += (k - banknotesUsed + 1) * denomination[i];
                  break;
              }
              minBurles += canUse * denomination[i];
              banknotesUsed += canUse;
          }
          out.println(minBurles);
      }
  
  
      public static FastReader sc;
      public static PrintWriter out;
      static class FastReader
      {
          BufferedReader br;
          StringTokenizer str;
  
          public FastReader()
          {
              br = new BufferedReader(new
                      InputStreamReader(System.in));
          }
  
          String next()
          {
              while (str == null || !str.hasMoreElements())
              {
                  try
                  {
                      str = new StringTokenizer(br.readLine());
                  }
                  catch (IOException  end)
                  {
                      end.printStackTrace();
                  }
              }
              return str.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 end)
              {
                  end.printStackTrace();
              }
              return str;
          }
      }
  }