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.*;
  import java.io.*;
  
  public class _116 {
      public static void main(String[] args) {
          MyScanner sc = new MyScanner();
          PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
          int t = sc.nextInt();
          long [] pow = new long[19];
          long start = 1;
          for (int i = 0; i < 19; i++) {
              pow[i] = start;
              start *= 10;
          }
          while (t-- > 0) {
              int n = sc.nextInt();
              long k = sc.nextLong();
              Set<Integer> set = new HashSet<>();
              for (int i = 0; i < n; i++) set.add(sc.nextInt());
              set.add(18);
              ArrayDeque<Long> res = new ArrayDeque<>();
              int cur = 0;
              long moves = 0;
              while (moves < k + 1) {
                  int s = cur;
                  cur++;
                  while (!set.contains(cur)) ++cur;
                  long p = pow[cur - s] - 1;
                  if (moves + p > k + 1) {
                      res.addFirst(k + 1 - moves);
                      moves = k + 1;
                  } else {
                      res.addFirst(p);
                      moves += p;
                  }
              }
              while (!res.isEmpty()) out.print(res.pollFirst());
              out.println();
          }
          out.close();
      }
  
  
      static void sort(int[] a) {
          ArrayList<Integer> q = new ArrayList<>();
          for (int i : a) q.add(i);
          Collections.sort(q);
          for (int i = 0; i < a.length; i++) a[i] = q.get(i);
      }
  
      static void sort(long[] a) {
          ArrayList<Long> q = new ArrayList<>();
          for (long i : a) q.add(i);
          Collections.sort(q);
          for (int i = 0; i < a.length; i++) a[i] = q.get(i);
      }
  
  
      //-----------MyScanner class for faster input----------
      public static class MyScanner {
          BufferedReader br;
          StringTokenizer st;
  
          public MyScanner() {
              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;
          }
  
  
      }
  
  }