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.InputStream;
  import java.io.InputStreamReader;
  import java.io.PrintWriter;
  import java.util.StringTokenizer;
  
  
  public class TaskC {
  
      public static void main(String[] arg) {
          final FastScanner in = new FastScanner(System.in);
          final PrintWriter out = new PrintWriter(System.out);
          final int T = in.nextInt();
          for (int t = 0; t < T; t++) {
              final int n = in.nextInt();
              final int k = in.nextInt();
              final int[] a = in.readIntArr(n);
              out.println(solution(n, k, a));
          }
          out.flush();
          out.close();
          in.close();
      }
  
      private static long solution(final long n, final long k, final int[] a) {
          long sum = 0;
          long coins = 0;
          for (int i = 0; i < a.length; i++) {
              long cur = pow10(a[i]);
              if (i + 1 < a.length) {
                  long next = pow10(a[i + 1]);
                  long cnt = next / cur - 1;
                  long add = Math.min(cnt, k - coins + 1);
                  coins += add;
                  sum += add * cur;
              } else {
                  long add = k - coins + 1;
                  coins += add;
                  sum += add * cur;
              }
              if (coins > k) {
                  return sum;
              }
          }
  
  
          return sum;
      }
  
      final static long pow10(final int a) {
          long res = 1;
          for (int i = 0; i < a; i++) {
              res *= 10;
          }
          return res;
      }
  
      private static class FastScanner {
          BufferedReader br;
          StringTokenizer st;
  
          FastScanner(InputStream stream) {
              try {
                  br = new BufferedReader(new InputStreamReader(stream));
              } catch (Exception e) {
                  e.printStackTrace();
              }
          }
  
          String next() {
              while (st == null || !st.hasMoreTokens()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          int[] readIntArr(int n) {
              int[] result = new int[n];
              for (int i = 0; i < n; i++) {
                  result[i] = Integer.parseInt(next());
              }
              return result;
          }
  
          long[] readLongArr(int n) {
              long[] result = new long[n];
              for (int i = 0; i < n; i++) {
                  result[i] = Long.parseLong(next());
              }
              return result;
          }
  
  
          void close() {
              try {
                  br.close();
              } catch (IOException e) {
                  e.printStackTrace();
              }
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
      }
  
  
  }