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)  → ...
// from Kunal's solution, i added comments to check my own understanding
import java.util.Scanner;

public class A {

	static int dp[][] = new int[30001][500]; // 500 is arbitrary
	static int offset = 250; // offset needed to account for the fact that jumps increase (l+1) and decrease (l-1)
	// so set the offset to the middle of the array [500] 
	
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt(), d = sc.nextInt();
		for(int i = 0; i<dp.length; i++) {
			for(int j = 0; j< dp[0].length; j++) dp[i][j] = -1; // initialize to -1
		}
		int freq[] = new int[30001]; // record locations of the gems
		for(int i = 0; i< n; i++) freq[sc.nextInt()]++;
		int x = recur(freq, d, d, d);
		System.out.println(x);
	}
	public static int recur(int freq[], int p, int l, int d) {
		if(p>30000) return 0; // not possible, out of bounds
		if(dp[p][l - d + offset]!=-1) return dp[p][l - d + offset]; // return if answer already calculated
		int ans = freq[p]; // answer = number of gems at p
		ans += Math.max(recur(freq, p+l, l, d), recur(freq, p+l+1, l+1, d) ); // max of jumping l or l + 1
		if(l>1) ans = Math.max(ans, freq[p] + recur(freq,p+l-1,l-1, d)); // consider l - 1 iff l > 1 to avoid jumping of l = 0
		return dp[p][l - d + offset] = ans; // record answer
	}
}

										 				 			 	 						 	