Problem: The Shuseki Islands are an archipelago of 30001 small islands in the Yutampo Sea. The islands are evenly spaced along a line, numbered from 0 to 30000 from the west to the east. These islands are known to contain many treasures. There are n gems in the Shuseki Islands in total, and the i-th gem is located on island pi.

Mr. Kitayuta has just arrived at island 0. With his great jumping ability, he will repeatedly perform jumps between islands to the east according to the following process: 

  * First, he will jump from island 0 to island d. 
  * After that, he will continue jumping according to the following rule. Let l be the length of the previous jump, that is, if his previous jump was from island prev to island cur, let l = cur - prev. He will perform a jump of length l - 1, l or l + 1 to the east. That is, he will jump to island (cur + l - 1), (cur + l) or (cur + l + 1) (if they exist). The length of a jump must be positive, that is, he cannot perform a jump of length 0 when l = 1. If there is no valid destination, he will stop jumping. 



Mr. Kitayuta will collect the gems on the islands visited during the process. Find the maximum number of gems that he can collect.

Input

The first line of the input contains two space-separated integers n and d (1 ≤ n, d ≤ 30000), denoting the number of the gems in the Shuseki Islands and the length of the Mr. Kitayuta's first jump, respectively.

The next n lines describe the location of the gems. The i-th of them (1 ≤ i ≤ n) contains a integer pi (d ≤ p1 ≤ p2 ≤ ... ≤ pn ≤ 30000), denoting the number of the island that contains the i-th gem.

Output

Print the maximum number of gems that Mr. Kitayuta can collect.

Examples

Input

4 10
10
21
27
27


Output

3


Input

8 8
9
19
28
36
45
55
66
78


Output

6


Input

13 7
8
8
9
16
17
17
18
21
23
24
24
26
30


Output

4

Note

In the first sample, the optimal route is 0  →  10 (+1 gem)  →  19  →  27 (+2 gems)  → ...

In the second sample, the optimal route is 0  →  8  →  15  →  21 →  28 (+1 gem)  →  36 (+1 gem)  →  45 (+1 gem)  →  55 (+1 gem)  →  66 (+1 gem)  →  78 (+1 gem)  → ...

In the third sample, the optimal route is 0  →  7  →  13  →  18 (+1 gem)  →  24 (+2 gems)  →  30 (+1 gem)  → ...

Solution (in JAVA):

Solution:
import java.io.BufferedReader;
import java.io.File;
import java.io.FileInputStream;
import java.io.FileNotFoundException;
import java.io.FileOutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Iterator;
import java.util.Objects;
import java.util.StringTokenizer;

public class Main {

  public static void main(String[] args) throws IOException {
    IOScanner scanner;
    if (new File("data.in").exists()) {
      scanner = new IOScanner("data.in", null);
    } else {
      scanner = new IOScanner();
    }
    while (scanner.hasNext()) {
      new Solver(scanner).solve();
    }
    scanner.close();
  }
}

class IOScanner {
  private final BufferedReader reader;
  private final PrintWriter writer;
  private StringTokenizer stringTokenizer;

  public IOScanner() {
    this(System.in, System.out);
  }

  public IOScanner(String input, String output) throws FileNotFoundException {
    this(input == null ? null : new File(input), output == null ? null : new File(output));
  }

  public IOScanner(File input, File output) throws FileNotFoundException {
    this(input == null ? System.in : new FileInputStream(input), output == null ? System.out
        : new FileOutputStream(output));
  }

  public IOScanner(InputStream inputStream, OutputStream outputStream) {
    reader = new BufferedReader(new InputStreamReader(inputStream));
    writer = new PrintWriter(outputStream);
  }

  private StringTokenizer read() {
    while (stringTokenizer == null || !stringTokenizer.hasMoreTokens()) {
      try {
        stringTokenizer = new StringTokenizer(reader.readLine());
      } catch (Exception e) {
        return null;
      }
    }
    return stringTokenizer;
  }

  public boolean hasNext() {
    StringTokenizer tokenizer = read();
    return tokenizer != null && tokenizer.hasMoreTokens();
  }

  public int nextInt() {
    return Integer.parseInt(next());
  }

  public long nextLong() {
    return Long.parseLong(next());
  }

  public double nextDouble() {
    return Double.parseDouble(next());
  }

  public int[] nextInts(int n) {
    int[] input = new int[n];
    for (int i = 0; i < n; i++) {
      input[i] = nextInt();
    }
    return input;
  }

  public int[][] nextInts(int n, int m) {
    int[][] input = new int[n][];
    for (int i = 0; i < n; i++) {
      input[i] = nextInts(m);
    }
    return input;
  }

  public double[] nextDoubles(int n) {
    double[] input = new double[n];
    for (int i = 0; i < n; i++) {
      input[i] = nextDouble();
    }
    return input;
  }

  public String[] nexts(int n) {
    String[] input = new String[n];
    for (int i = 0; i < n; i++) {
      input[i] = next();
    }
    return input;
  }

  public char[][] nextCharArrays(int n) {
    String[] input = nexts(n);
    char[][] result = new char[n][];
    for (int i = 0; i < n; i++) {
      result[i] = input[i].toCharArray();
    }
    return result;
  }

  public String next() {
    return read().nextToken();
  }

  public void println() {
    writer.println();
  }

  public void println(Iterable<?> iterable) {
    println(iterable.iterator());
  }

  public void println(Iterator<?> iterator) {
    boolean first = true;
    while (iterator.hasNext()) {
      if (first) {
        first = false;
      } else {
        print(' ');
      }
      print(iterator.next());
    }
    println();
  }

  public void println(Object token) {
    writer.println(token);
  }

  public void print(Object token) {
    writer.print(token);
  }

  public void printf(String format, Object... args) {
    writer.printf(format, args);
  }

  public void flush() {
    writer.flush();
  }

  public void close() throws IOException {
    reader.close();
    writer.close();
  }
}

class Edge {
  final int from, to, type;
  boolean needed = false;
  final Edge next;
  final String s;

  Edge(int from, int to, int type, Edge next, String s) {
    this.from = from;
    this.to = to;
    this.type = type;
    this.next = next;
    this.s = s;
  }

  @Override
  public String toString() {
    return s;
  }

  @Override
  public boolean equals(Object other) {
    if (!(other instanceof Edge)) {
      return false;
    }
    Edge that = (Edge) other;
    return Objects.equals(this.s, that.s);
  }

  @Override
  public int hashCode() {
    return Objects.hash(s);
  }
}

class Solver {

  private static final int M = 30001;
  private final IOScanner scanner;
  private int bound;
  private int[][] dp;
  private int[] gem;
  private int m;

  public Solver(IOScanner scanner) {
    this.scanner = scanner;
  }

  public void solve() {
    int n = scanner.nextInt();
    int d = scanner.nextInt();
    gem = new int[M];
    Arrays.fill(gem, 0);
    for (int p : scanner.nextInts(n)) {
      gem[p]++;
      m = p;
    }
    dp = new int[M][505];
    for (int i = 0; i < M; i++) {
      Arrays.fill(dp[i], -1);
    }
    bound = d - 250;
    scanner.println(dfs(d, d));
  }

  private int dfs(int pos, int d) {
    if (pos > m) {
      return 0;
    }
    if (dp[pos][d - bound] >= 0) {
      return dp[pos][d - bound];
    }
    int max = 0;
    for (int nd = Math.max(1, d - 1); nd <= d + 1; nd++) {
      max = Math.max(max, dfs(pos + nd, nd));
    }
    dp[pos][d - bound] = max + gem[pos];
    return dp[pos][d - bound];
  }
}