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:
/**
 * DA-IICT
 * Author : PARTH PATEL
 */
import java.io.*;
import java.math.*;
import java.util.*;

import static java.util.Arrays.fill;
import static java.lang.Math.*;
import static java.util.Arrays.sort;
import static java.util.Collections.sort;


public class C505 
{

	public static int mod = 1000000007;
	static FasterScanner in = new FasterScanner();
	static PrintWriter out = new PrintWriter(System.out);
	static int n,d;
	static int[][] dp;
	static int[] position;
	public static int func(int index,int lastjump)
	{
		if(index>30000 || lastjump<=0)
			return 0;
		if(dp[index][lastjump-d+255]!=-1)
			return dp[index][lastjump-d+255];
		int answer=-1;
		answer=func(index+lastjump, lastjump);
		answer=max(answer, func(index+lastjump+1, lastjump+1));
		answer=max(answer, func(index+lastjump-1, lastjump-1));
		answer+=position[index];
		return dp[index][lastjump-d+255]=answer;
	}
	public static void main(String[] args) 
	{
		n=in.nextInt();
		d=in.nextInt();
		dp=new int[30001][550];
		position=new int[30001];
		for(int i=0;i<=30000;i++)
		{
			for(int j=0;j<550;j++)
			{
				dp[i][j]=-1;
			}
		}
		for(int i=1;i<=n;i++)
		{
			position[in.nextInt()]++;
		}
		out.println(func(d, d));
		out.close();

	}
	

	/////////////////SEGMENT TREE (BUILD-UPDATE-QUERY)/////////////////////////////
	/////////////////UPDATE FOLLOWING METHODS AS PER NEED//////////////////////////
	/*
	public static void buildsegmenttree(int node,int start,int end)
	{
		if(start==end)
		{
			// Leaf node will have a single element
			segmenttree[node]=arr[start];
		}
		else
		{
			int mid=start+(end-start)/2;
			// Recurse on the left child
			buildsegmenttree(2*node, start, mid);
			// Recurse on the right child
			buildsegmenttree(2*node+1, mid+1, end);
			// Internal node will have the sum of both of its children
			segmenttree[node]=segmenttree[2*node]+segmenttree[2*node+1];
		}
	}
	
	public static void updatesegmenttree(int node,int start,int end,int idx,int val)
	{
		if(start==end)
		{
			//Leaf Node
			arr[idx]+=val;
			segmenttree[node]+=val;
		}
		else
		{
			int mid=start+(end-start)/2;
			if(start<=idx && idx<=mid)
			{
				// If idx is in the left child, recurse on the left child
				updatesegmenttree(2*node, start, mid, idx, val);
			}
			else
			{
				// if idx is in the right child, recurse on the right child
				updatesegmenttree(2*node+1, mid+1, end, idx, val);
			}
			// Internal node will have the sum of both of its children
			segmenttree[node]=segmenttree[2*node]+segmenttree[2*node+1];
		}
	}
	
	public static long querysegmenttree(int node,int start,int end,int l,int r)
	{
		if(r<start || end<l)
		{
			// range represented by a node is completely outside the given range
			return 0;
		}
		if(l <= start && end <= r)
	    {
	        // range represented by a node is completely inside the given range
	        return segmenttree[node];
	    }
		// range represented by a node is partially inside and partially outside the given range
		int mid=start+(end-start)/2;
		long leftchild=querysegmenttree(2*node, start, mid, l, r);
		long rightchild=querysegmenttree(2*node+1, mid+1, end, l, r);
		return (leftchild+rightchild);
	}
	*/
	
	public static long pow(long x, long n, long mod) 
	{
		long res = 1;
		for (long p = x; n > 0; n >>= 1, p = (p * p) % mod) 
		{
			if ((n & 1) != 0) 
			{
				res = (res * p % mod);
			}
		}
		return res;
	}

	public static long gcd(long n1, long n2) 
	{
		long r;
		while (n2 != 0) 
		{
			r = n1 % n2;
			n1 = n2;
			n2 = r;
		}
		return n1;
	}

	public static long lcm(long n1, long n2) 
	{
		long answer = (n1 * n2) / (gcd(n1, n2));
		return answer;
	}

	static class FasterScanner 
	{
		private byte[] buf = new byte[1024];
		private int curChar;
		private int snumChars;

		public int read() 
		{
			if (snumChars == -1)
				throw new InputMismatchException();
			if (curChar >= snumChars) 
			{
				curChar = 0;
				try 
				{
					snumChars = System.in.read(buf);
				} catch (IOException e) 
				{
					throw new InputMismatchException();
				}
				if (snumChars <= 0)
					return -1;
			}
			return buf[curChar++];
		}

		public String nextLine() 
		{
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do 
			{
				res.appendCodePoint(c);
				c = read();
			} while (!isEndOfLine(c));
			return res.toString();
		}

		public String nextString() 
		{
			int c = read();
			while (isSpaceChar(c))
				c = read();
			StringBuilder res = new StringBuilder();
			do 
			{
				res.appendCodePoint(c);
				c = read();
			} while (!isSpaceChar(c));
			return res.toString();
		}

		public long nextLong() 
		{
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-') 
			{
				sgn = -1;
				c = read();
			}
			long res = 0;
			do 
			{
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			} 
			while (!isSpaceChar(c));
			return res * sgn;
		}

		public int nextInt() 
		{
			int c = read();
			while (isSpaceChar(c))
				c = read();
			int sgn = 1;
			if (c == '-') 
			{
				sgn = -1;
				c = read();
			}
			int res = 0;
			do 
			{
				if (c < '0' || c > '9')
					throw new InputMismatchException();
				res *= 10;
				res += c - '0';
				c = read();
			} 
			while (!isSpaceChar(c));
			return res * sgn;
		}

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

		public long[] nextLongArray(int n) 
		{
			long[] arr = new long[n];
			for (int i = 0; i < n; i++)
			{
				arr[i] = nextLong();
			}
			return arr;
		}

		private boolean isSpaceChar(int c) 
		{
			return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
		}

		private boolean isEndOfLine(int c) 
		{
			return c == '\n' || c == '\r' || c == -1;
		}
	}

}