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 Main {
      public static void main(String[] args) throws IOException {
          //br = new BufferedReader(new FileReader("input.txt"));
          //out = new PrintWriter("output.txt");
          int t = 1;
          t = nextInt();
          while (t-- != 0) {
              int n = nextInt();
              long k = nextLong();
              long ans = 0;
              long[] a = new long[n];
              for (int i = 0; i < n; i++) {
                  a[i] = nextInt();
              }
              long q = 0;
              for (int i = 0; i < a.length - 1; i++) {
                  long y = a[i + 1] - a[i];
                  long u = 0;
                  for (int j = 0; j < y; j++) {
                      u *= 10;
                      u += 9;
                  }
                  if (q + u > k || i == a.length - 2) {
                      if (q + u > k) {
                          if (ans == 0) ans = k - q + 1;
                          else ans = Long.parseLong((k - q + 1) + "" + ans);
                      } else {
                          if (ans == 0) ans = Long.parseLong((k - q - u + 1) + "" + u);
                          else ans = Long.parseLong((k - q - u + 1) + "" + u + "" + ans);
                      }
                      break;
                  } else {
                      k -= u;
                      if (ans == 0) ans = u;
                      else ans = Long.parseLong(u + "" + ans);
                  }
              }
              if (ans == 0) {
                  out.println(k + 1);
              } else out.println(ans);
          }
          out.close();
      }
  
      static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
      static PrintWriter out = new PrintWriter(System.out);
      static StringTokenizer in = new StringTokenizer("");
  
  
      public static boolean hasNext() throws IOException {
          if (in.hasMoreTokens()) return true;
          String s;
          while ((s = br.readLine()) != null) {
              in = new StringTokenizer(s);
              if (in.hasMoreTokens()) return true;
          }
          return false;
      }
  
      public static String nextToken() throws IOException {
          while (!in.hasMoreTokens()) {
              in = new StringTokenizer(br.readLine());
          }
          return in.nextToken();
      }
  
      public static int nextInt() throws IOException {
          return Integer.parseInt(nextToken());
      }
  
      public static long nextLong() throws IOException {
          return Long.parseLong(nextToken());
      }
  
      public static double nextDouble() throws IOException {
          return Double.parseDouble(nextToken());
      }
  }