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.*;
import java.util.*;


public class Main {

  public static void main(String[] args) {
    solve();
  }

  final static int MAXN = 30005;
  final static int MAXK = 505;
  static int N = 0;
  static int M = 0;
  static int K = 0;
  static int W = 0;
  static int H = 0;
  static int[] A;
  static long[][] dp;
  static int[][] pre;
  static int[][] g;
  static int curAns = Integer.MAX_VALUE;
  static long[] HASH = new long[MAXN];
  static long MOD = (1L << 50) + 7;
  static int[][] memo;
  static boolean[][] seen;

  static void solve() {
    FastReader scanner = new FastReader();
    PrintWriter out = new PrintWriter(System.out);
    N = scanner.nextInt();
    M = scanner.nextInt();
    A = new int[MAXN];
    for (int i = 0; i < N; i++) {
      int v = scanner.nextInt();
      A[v] += 1;
    }

    memo = new int[MAXN][500];
    seen = new boolean[MAXN][500];

    int ans = dfs(M, M);
    out.println(ans);
    out.close();
  }



  static int dfs(int pos, int jump) {
    if (pos >= MAXN || jump <= 0) {
      return 0;
    }

    int key = jump - (M - 250);

    if (seen[pos][key]) {
      return memo[pos][key];
    }

    int ans = A[pos] + Math.max(Math.max(dfs(pos+jump, jump), dfs(pos+jump-1, jump-1)), dfs(pos+jump+1, jump+1));
    seen[pos][key] = true;
    memo[pos][key] = ans;

    return ans;
  }


  public static class FastReader {
    BufferedReader br;
    StringTokenizer st;


    public FastReader() {
      br = new BufferedReader(new InputStreamReader(System.in));
    }

    String next() {
      while (st == null || !st.hasMoreTokens()) {
        try {
          st = new StringTokenizer(br.readLine());
        } catch (Exception r) {
          r.printStackTrace();
        }
      }
      return st.nextToken();
    }

    int nextInt() {
      return Integer.parseInt(next());//converts string to integer
    }

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

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

    String nextLine() {
      String str = "";
      try {
        str = br.readLine();
      } catch (Exception r) {
        r.printStackTrace();
      }
      return str;
    }
  }
}

class UnioinSet {
  private Map<Long, Long> parents = new HashMap<>();

  public Long find(Long u) {
    if (parents.getOrDefault(u, u).equals(u)) {
      return u;
    }

    return find(parents.get(u));
  }

  public void union(Long u, Long v) {
    Long pu = find(u);
    Long pv = find(v);

    parents.put(pu, pv);
  }
}


class Pair<A, B> implements Comparable<Pair<A, B>>{
  private A first;
  private B second;

  public Pair(A first, B second) {
    super();
    this.first = first;
    this.second = second;
  }

  public String toString()
  {
    return "(" + first + ", " + second + ")";
  }

  public A getFirst() {
    return first;
  }

  public void setFirst(A first) {
    this.first = first;
  }

  public B getSecond() {
    return second;
  }

  public void setSecond(B second) {
    this.second = second;
  }

  @Override
  public int compareTo(Pair<A, B> o) {
    int cmp = compare(first, o.first);
    return cmp == 0 ? compare(second, o.second) : cmp;
  }

  private static int compare(Object o1, Object o2) {
    return o1 == null ? o2 == null ? 0 : -1 : o2 == null ? +1
      : ((Comparable) o1).compareTo(o2);
  }

  @Override
  public int hashCode() {
    return 31 * hashcode(first) + hashcode(second);
  }

  private static int hashcode(Object o) {
    return o == null ? 0 : o.hashCode();
  }

  public boolean equals(Object other) {
    if (other instanceof Pair) {
      Pair<A, B> otherPair = (Pair<A, B>) other;
      return
        ((  this.first == otherPair.first ||
          ( this.first != null && otherPair.first != null &&
            this.first.equals(otherPair.first))) &&
          (  this.second == otherPair.second ||
            ( this.second != null && otherPair.second != null &&
              this.second.equals(otherPair.second))) );
    }

    return false;
  }
}