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.Arrays;
  import java.util.Random;
  import java.util.StringTokenizer;
  
  public class codeforces_Edu116_C {
      private static void solve(FastIOAdapter io) {
          int n = io.nextInt();
          int k = io.nextInt();
          int[] a = io.readArray(n);
  
          long ans = 0;
          int left = k + 1;
          for (int i = 0; i < n && left > 0; i++) {
              int possibleDiff;
              if (i < n - 1) {
                  possibleDiff = (int)Math.pow(10, a[i + 1])/(int)Math.pow(10, a[i]) - 1;
              } else {
                  possibleDiff = Integer.MAX_VALUE;
              }
              int diff = Math.min(left, possibleDiff);
              ans += (long) Math.pow(10, a[i]) * diff;
              left -= diff;
          }
          io.out.println(ans);
      }
  
      public static void main(String[] args) throws Exception {
          try (FastIOAdapter ioAdapter = new FastIOAdapter()) {
              int count = 1;
              count = ioAdapter.nextInt();
              while (count-- > 0) {
                  solve(ioAdapter);
              }
          }
      }
  
      static void ruffleSort(int[] arr) {
          int n = arr.length;
          Random rnd = new Random();
          for (int i = 0; i < n; ++i) {
              int tmp = arr[i];
              int randomPos = i + rnd.nextInt(n - i);
              arr[i] = arr[randomPos];
              arr[randomPos] = tmp;
          }
          Arrays.sort(arr);
      }
  
      static class FastIOAdapter implements AutoCloseable {
          BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
          public PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter((System.out))));
          StringTokenizer st = new StringTokenizer("");
  
          String next() {
              while (!st.hasMoreTokens())
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              return st.nextToken();
          }
  
          String nextLine() {
              try {
                  return br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
                  return null;
              }
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          int[] readArray(int n) {
              int[] a = new int[n];
              for (int i = 0; i < n; i++) a[i] = nextInt();
              return a;
          }
  
          long[] readArrayLong(int n) {
              long[] a = new long[n];
              for (int i = 0; i < n; i++) a[i] = nextLong();
              return a;
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          @Override
          public void close() throws Exception {
              out.flush();
              out.close();
              br.close();
          }
      }
  }
  