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:
/** https://codeforces.com/problemset/problem/505/C
 * #bfs #dp
 */
import java.util.Scanner;
import java.util.Queue;
import java.util.LinkedList;
import java.util.Arrays;
import java.util.StringTokenizer;
import java.io.*;
/* Bottom Up - take time, rework
public class TreasureHunter {
    private static class Node {
        int idx, jump;
        Node(int idx, int jump) {
            this.idx = idx;
            this.jump = jump;
        }
    }
    public static void main(String[] args) {
        MyScanner sc = new MyScanner(System.in);
        int max = 30001;
        int n = sc.nextInt();
        int d = sc.nextInt();
        int[] arr = new int[max];
        int last_pos = 0;
        for (int i = 0; i < n; i++){
            last_pos = sc.nextInt();
            arr[last_pos] += 1;
        }
        max = last_pos + 1;
        int max_jump = d + n;
/* Find max position that can be reached.
        // d = 1
        // 0, d, d+1, d+2, d+3, ... >= MAX_N
        // d + d+1 + d+2 + ... + d+k >= MAX_N
        // (k+1)*d + k*(k+1)/2 >= MAX_N
        // k+1 + k(k+1)/2 >= MAX_N
        // k^2 + 3k + 2 => 2 * MAX_N = 60000
        // k(k + 1) + 2(k + 1)
        // (k+1) * (k+2) >= 60000
        // => k = 245
        
        // min_jump = d - 245 => 0
        // max_jump = d + 245 => 500
        => max_jump = 500
 *//*
        max_jump = 500;
        int[][] gems = new int[max][max_jump];
        for(int[] gems_pos : gems) {
            Arrays.fill(gems_pos, -1);
        }

        int result = gems[d][d % max_jump] = arr[d];
        int[] slices = {-1, 0, 1};
        Queue<Node> q = new LinkedList<>();
        q.add(new Node(d, d));
        while(!q.isEmpty()) {
            Node node = q.poll();
            for (int slice : slices) {
                int jump = slice + node.jump;
                int next = node.idx + jump;
                if (jump > 0 && next < max) {
                    if (arr[next] + gems[node.idx][node.jump % max_jump] > gems[next][jump % max_jump]) {
                        gems[next][jump % max_jump] = arr[next] + gems[node.idx][node.jump % max_jump];
                        if (result < gems[next][jump % max_jump]) {
                            result = gems[next][jump % max_jump];
                        }
                        q.add(new Node(next, jump));
                    }
                }
            }
        }

        System.out.println(result);
    }
}

class MyScanner {
    BufferedReader br;
    StringTokenizer st;

    public MyScanner(InputStream is) {
        br = new BufferedReader(new InputStreamReader(is));
    }
    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;
    }
}

*/
// Top down
/**
/*
dp[i][j]: max gem from n to i, last jump j
    -> dp[i+j-1][j-1]   if i+j-1 <= n
    -> dp[i+j  ][j  ]   if i+j   <= n
    -> dp[i+j+1][j+1]   if i+j+1 <= n
    dp(i, j) = arr[i] + max(dp(i+j-1, j-1), dp(i+j, j), dp(i+j+1, j+1))

dp(d, d)

DP top-down
function dp(int i, int j):
  if f[i][j] != -1:
    return f[i][j]
  if i == n:
    return f[i][j] = arr[i]
  f[i][j] = 0
  if step j-1 is valid:
    f[i][j] = max(f[i][j], arr[i] + dp(i+j-1, j-1))
  if step j is valid:
    f[i][j] = max(f[i][j], arr[i] + dp(i+j, j))
  if step j-1 is valid:
    f[i][j] = max(f[i][j], arr[i] + dp(i+j+1, j+1))
  return f[i][j]
*/
public class TreasureHunter {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int max = 30001;
        int n = sc.nextInt();
        int d = sc.nextInt();
        int[] arr = new int[max];
        int last_pos = 0;
        for (int i = 0; i < n; i++){
            last_pos = sc.nextInt();
            arr[last_pos] += 1;
        }
        int max_jump = 500;
        int[][] gems = new int[last_pos+1][max_jump];
        for(int[] gems_pos : gems) {
            Arrays.fill(gems_pos, -1);
        }
        int result = dp(d, d, d, gems, arr);
        System.out.println(result);
    }
    // d + 250                  d           <=>     250             (d-250)
    // j                       j-d+250      
    // -250
    private static int dp(int i, int j, int d, int[][] gems, int[] arr) {
        int n = gems.length;
        int jump = j - d + 250;
        if (gems[i][jump] != -1) {
            return gems[i][jump];
        }
        if (i == n - 1) {
            gems[i][jump] = arr[i];
            return gems[i][jump];
        }
        // initialize
        gems[i][jump] = 0;

        // j - 1
        if (i + j - 1 < n && j - 1 > 0 && Math.abs(j - d) < 250) {
            gems[i][jump] = Math.max(gems[i][jump], dp(i + j - 1, j - 1, d, gems, arr));
        }
        // j
        if (i + j < n && j > 0 && Math.abs(j - d) < 250) {
            gems[i][jump] = Math.max(gems[i][jump], dp(i + j, j, d, gems, arr));
        }
        // j + 1
        if (i + j + 1 < n && j + 1 > 0 && Math.abs(j - d) < 250) {
            gems[i][jump] = Math.max(gems[i][jump], dp(i + j + 1, j + 1, d, gems, arr));
        }

        gems[i][jump] += arr[i];

        return gems[i][jump];
    }
}