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.BufferedInputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.text.DecimalFormat;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.InputMismatchException;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.Map.Entry;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Random;
import java.util.Set;
import java.util.SortedSet;
import java.util.Stack;
import java.util.TreeMap;
import java.util.TreeSet;

import javax.net.ssl.SSLContext;

public class Main
{	
	/********************************************** a list of common variables **********************************************/
	private MyScanner scan = new MyScanner();
	private PrintWriter out = new PrintWriter(System.out);
	private final double PI = Math.acos(-1.0);
	private final double EPS = 1e-6;
	private final int SIZEN = (int)(1e5);
	private final int MOD = (int)(1e9 + 7);
	private final long MODH = 10000000007L, BASE = 10007;
	private final int[] DX = {0, 1, 0, -1}, DY = {-1, 0, 1, 0};
	private ArrayList<Integer>[] edge;

  	public void foo()
  	{
  		int n = scan.nextInt();
  		int d = scan.nextInt();
  		final int M = 30000, S = 250;
  		int[] x = new int[M + 1];
  		for(int i = 0;i < n;++i) x[scan.nextInt()] += 1;
  		int[][] dp = new int[M + 1][2 * S + 1];
  		dp[d][S] = 1 + x[d];
  		int ans = 1 + x[d];
  		for(int i = d;i <= M;++i)
  		{
  			for(int j = 0;j <= 2 * S;++j)
  			{
  				if(dp[i][j] > 0)
  				{
  					int f = d + j - S;
  					if(f > 1 && i + f - 1 <= M && dp[i + f - 1][j - 1] < dp[i][j] + x[i + f - 1])
  					{
  						dp[i + f - 1][j - 1] = dp[i][j] + x[i + f - 1];
  						ans = Math.max(ans, dp[i + f - 1][j - 1]);
  					}
  					if(i + f <= M && dp[i + f][j] < dp[i][j] + x[i + f])
  					{
  						dp[i + f][j] = dp[i][j] + x[i + f];
  						ans = Math.max(ans, dp[i + f][j]);
  					}
  					if(i + f + 1 <= M && dp[i + f + 1][j + 1] < dp[i][j] + x[i + f + 1])
  					{
  						dp[i + f + 1][j + 1] = dp[i][j] + x[i + f + 1];
  						ans = Math.max(ans, dp[i + f + 1][j + 1]);
  					}
  				}
  			}
  		}
  		out.println(ans - 1);
  	}

	public static void main(String[] args)
	{
		Main m = new Main();
		m.foo();
		m.out.close();
	}

	/********************************************** a list of common algorithms **********************************************/
	/**
	 * 1---Get greatest common divisor
	 * @param a :	first number
	 * @param b :	second number
	 * @return		greatest common divisor
	 */
	public long gcd(long a, long b)
	{
		return 0 == b ? a : gcd(b, a % b);
	}
	
	/**
	 * 2---Get the distance from a point to a line
	 * @param x1	the x coordinate of one endpoint of the line
	 * @param y1	the y coordinate of one endpoint of the line
	 * @param x2	the x coordinate of the other endpoint of the line
	 * @param y2	the y coordinate of the other endpoint of the line
	 * @param x		the x coordinate of the point 
	 * @param y		the x coordinate of the point
	 * @return		the distance from a point to a line
	 */
	public double getDist(long x1, long y1, long x2, long y2, long x, long y)
	{
		long a = y2 - y1;
		long b = x1 - x2;
		long c = y1 * (x2 - x1) - x1 * (y2 - y1);
		return Math.abs(a * x + b * y + c) / Math.sqrt(a * a + b * b);
	}
	
	/**
	 * 3---Get the distance from one point to a segment (not a line)
	 * @param x1	the x coordinate of one endpoint of the segment
	 * @param y1	the y coordinate of one endpoint of the segment
	 * @param x2	the x coordinate of the other endpoint of the segment
	 * @param y2	the y coordinate of the other endpoint of the segment
	 * @param x		the x coordinate of the point
	 * @param y		the y coordinate of the point
	 * @return		the distance from one point to a segment (not a line)
	 */
	public double ptToSeg(long x1, long y1, long x2, long y2, long x, long y)
	{
		double cross = (x2 - x1) * (x - x1) + (y2 - y1) * (y - y1);
		if(cross <= 0)
		{
			return (x - x1) * (x - x1) + (y - y1) * (y - y1);
		}
		double d = (x2 - x1) * (x2 - x1) + (y2 - y1) * (y2 - y1);
		if(cross >= d) 
		{
			return (x - x2) * (x - x2) + (y - y2) * (y - y2);
		}
		double r = cross / d;
		double px = x1 + (x2 - x1) * r;
		double py = y1 + (y2 - y1) * r;
		return (x - px) * (x - px) + (y - py) * (y - py);
	}
	
	/**
	 * 4---KMP match, i.e. kmpMatch("abcd", "bcd") = 1, kmpMatch("abcd", "bfcd") = -1.
	 * @param t:	String to match.
	 * @param p:	String to be matched.
	 * @return		if can match, first index; otherwise -1.
	 */
	public int kmpMatch(char[] t, char[] p)
	{
		int n = t.length;
		int m = p.length;
		int[] next = new int[m + 1];
		next[0] = -1;
		int j = -1;
		for(int i = 1;i < m;++i)
		{
			while(j >= 0 && p[i] != p[j + 1])
			{
				j = next[j];
			}
			if(p[i] == p[j + 1])
			{
				++j;
			}
			next[i] = j;
		}
		j = -1;
		for(int i = 0;i < n;++i)
		{
			while(j >= 0 && t[i] != p[j + 1])
			{
				j = next[j];
			}
			if(t[i] == p[j + 1])
			{
				++j;
			}
			if(j == m - 1)
			{
				return i - m + 1;
			}
		}
		return -1;
	}
	
	/**
	 * 5---Get the hash code of a String
	 * @param	s:	input string
	 * @return	hash code
	 */
	public long hash(String s)
	{
		long key = 0, t = 1;
		for(int i = 0;i < s.length();++i)
		{
			key = (key + s.charAt(i) * t) % MODH;
			t = t * BASE % MODH;
		}
		return key;
	}
	
	/**
	 * 6---Get x ^ n % MOD quickly.
	 * @param	x:	base
	 * @param 	n:	times
	 * @return	x ^ n % MOD
	 */
	public long quickMod(long x, long n)
	{
		long ans = 1;
		while(n > 0)
		{
			if(1 == n % 2)
			{
				ans = ans * x % MOD;
 			}
			x = x * x % MOD;
			n >>= 1;
		}
		return ans;
	}
	
	/**
	 * 7---judge if a point is located inside a polygon
	 * @param x0	the x coordinate of the point
	 * @param y0	the y coordinate of the point
	 * @return true if it is inside the polygon, otherwise false
	 */
	/*public boolean contains(double x0, double y0)
	{
		int cross = 0;
		for(int i = 0;i < n;++i)
		{
			double s = x[i + 1] == x[i] ? 100000000 : (double)(y[i + 1] - y[i]) / (x[i + 1] - x[i]);
			boolean b1 = x[i] <= x0 && x0 < x[i + 1];
			boolean b2 = x[i + 1] <= x0 && x0 < x[i];
			boolean b3 = y0 < s * (x0 - x[i]) + y[i];
			if((b1 || b2) && b3) ++cross;
		}
		return cross % 2 != 0;
	}*/
	
	/**
	 * 8---judge if a point is located on the segment
	 * @param	x1	the x coordinate of one point of the segment
	 * @param	y1	the y coordinate of one point of the segment
	 * @param 	x2	the x coordinate of another point of the segment
	 * @param 	y2	the y coordinate of another point of the segment
	 * @param	x	the x coordinate of the point
	 * @param 	y	the y coordinate of the point
	 * @return	true if it is located on the segment, otherwise false
	 */
	public boolean isOnSeg(long x1, long y1, long x2, long y2, long x, long y)
	{
		return (x - x1) * (y2 - y1) == (x2 - x1) * (y - y1) && 
				x >= Math.min(x1, x2) && x <= Math.max(x1, x2) && 
				y >= Math.min(y1, y2) && y <= Math.max(y1, y2);
	}
	
	/**
	 * 9---get the cross product
	 * @param 	p1	point A
	 * @param 	p2	point B
	 * @param 	p	point O
	 * @return	cross product of OA x OB
	 */
	/*public long cross(Point p1, Point p2, Point p)
	{
		return (long)(p1.x - p.x) * (p2.y - p.y) - (long)(p2.x - p.x) * (p1.y - p.y);
	}*/
	
	class MyScanner
	{
	    private byte[] buf = new byte[1024];
	    private int curChar;
	    private int numChars;
	    BufferedInputStream bis = new BufferedInputStream(System.in);

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

	    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 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 double nextDouble() 
	    {
	        int c = read();
	        while (isSpaceChar(c))
	        {
	        	c = read();
	        }
	        int sgn = 1;
	        if (c == '-')
	        {
	            sgn = -1;
	            c = read();
	        }
	        double res = 0;
	        while (!isSpaceChar(c) && c != '.') 
	        {
	            if (c == 'e' || c == 'E')
	            {
	            	return res * Math.pow(10, nextInt());
	            }
	            if (c < '0' || c > '9')
	            {
	            	throw new InputMismatchException();
	            }
	            res *= 10;
	            res += c & 15;
	            c = read();
	        }
	        if (c == '.') 
	        {
	            c = read();
	            double m = 1;
	            while (!isSpaceChar(c)) 
	            {
	                if (c == 'e' || c == 'E')
	                {
	                	return res * Math.pow(10, nextInt());
	                }
	                if (c < '0' || c > '9')
	                {
	                	throw new InputMismatchException();
	                }
	                m /= 10;
	                res += (c & 15) * m;
	                c = read();
	            }
	        }
	        return res * sgn;
	    }
	    
	    public String next()
	    {
	        int c = read();
	        while (isSpaceChar(c))
	        {
	        	c = read();
	        }
	        StringBuilder res = new StringBuilder();
	        do 
	        {
	            res.appendCodePoint(c);
	            c = read();
	        } while (!isSpaceChar(c));
	        return res.toString();
	    }

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