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.util.*;
import java.io.*;

public class a implements Runnable {
    static Map<Integer,Integer> map;

    static int f(int curr,int d,int ord,int [][] dp){
        if(curr>=30001)return 0;
        int dd=d-(ord-250);
        if(dp[curr][dd]!=-1)return dp[curr][dd];

        int ans=0;
        for(int i=Math.max(d-1,1);i<=Math.min(d+1,30000);i++){
            int cnt=f(curr+i,i,ord,dp);
            cnt+=map.getOrDefault(curr+i, 0);
            ans=Math.max(ans,cnt);
        }

        return dp[curr][dd]=ans;
    }


    public static void main(String [] sst) throws IOException{
      Thread th=new Thread(null,new a(),"name",1<<24);   
      th.start();
    }

    public void run(){
        FastScanner sc=new FastScanner();
        int n=0,d=0;
        try{
            n=sc.nextInt();
            d=sc.nextInt();
            map=new HashMap<>();
            for(int i=0;i<n;i++){
                int x=sc.nextInt();
                map.put(x,map.getOrDefault(x, 0)+1);
            }
    }catch(IOException e){

    }
        int [][] dp=new int[30001][500];
        for(int i=0;i<dp.length;i++)Arrays.fill(dp[i],-1);
        int ans=f(d,d,d,dp);
        ans+=map.getOrDefault(d,0);
        System.out.println(ans);
    }

}







// 612-3706163

// sbi.03129@sbi.co.in





























class FastScanner{	
    private int BUFFER_SIZE = 1 << 16;
   
   private  DataInputStream din;
   
   private  byte[] buffer;
   
   private int bufferPointer, bytesRead;
   
   public FastScanner() {
       din = new DataInputStream(System.in);
       buffer = new byte[BUFFER_SIZE];
       bufferPointer = bytesRead = 0;
   }

   public FastScanner( String file_name) throws IOException {
       din = new DataInputStream(new FileInputStream(file_name));
       buffer = new byte[BUFFER_SIZE];
       bufferPointer = bytesRead = 0;
   }
   
   public String readLine() throws IOException {
        byte[] buf = new byte[64];
       int cnt = 0, c;
       while ((c = read()) != -1) {
           if (c == '\n')
               break;
           buf[cnt++] = (byte) c;
       }
       return new String(buf, 0, cnt);
   }
   
   public String next() throws IOException{

       byte c = read();
       while(Character.isWhitespace(c)){
           c = read();
       }
       
       StringBuilder builder = new StringBuilder();
       builder.append((char)c);
       c = read();
       while(!Character.isWhitespace(c)){
           builder.append((char)c);
           c = read();
       }
       
       return builder.toString();
   }

   public int nextInt() throws IOException {
       int ret = 0;
       byte c = read();
       while (c <= ' ')
           c = read();
        boolean neg = (c == '-');
       if (neg)
           c = read();
       do {
           ret = ret * 10 + c - '0';
       } while ((c = read()) >= '0' && c <= '9');

       if (neg)
           return -ret;
       return ret;
   }
   
   public int[] nextIntArray( int n) throws IOException {
        int arr[] = new int[n];
       for(int i = 0; i < n; i++){
           arr[i] = nextInt();
       }
       return arr;
   }

   public long nextLong() throws IOException {
       long ret = 0;
       byte c = read();
       while (c <= ' ')
           c = read();
        boolean neg = (c == '-');
       if (neg)
           c = read();
       do {
           ret = ret * 10 + c - '0';
       } while ((c = read()) >= '0' && c <= '9');
       if (neg)
           return -ret;
       return ret;
   }
   
   public long[] nextLongArray( int n) throws IOException {
        long arr[] = new long[n];
       for(int i = 0; i < n; i++){
           arr[i] = nextLong();
       }
       return arr;
   }

   public char nextChar() throws IOException{
       byte c = read();
       while(Character.isWhitespace(c)){
           c = read();
       }
       return (char) c;	
   }
   
   public double nextDouble() throws IOException {
       double ret = 0, div = 1;
       byte c = read();
       while (c <= ' ')
           c = read();
        boolean neg = (c == '-');
       if (neg)
           c = read();

       do {
           ret = ret * 10 + c - '0';
       } while ((c = read()) >= '0' && c <= '9');

       if (c == '.') {
           while ((c = read()) >= '0' && c <= '9') {
               ret += (c - '0') / (div *= 10);
           }
       }

       if (neg)
           return -ret;
       return ret;
   }
   
   public double[] nextDoubleArray( int n) throws IOException {
        double arr[] = new double[n];
       for(int i = 0; i < n; i++){
           arr[i] = nextDouble();
       }
       return arr;
   }

   private void fillBuffer() throws IOException {
       bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
       if (bytesRead == -1)
           buffer[0] = -1;
   }

   private byte read() throws IOException {
       if (bufferPointer == bytesRead)
           fillBuffer();
       return buffer[bufferPointer++];
   }

   public void close() throws IOException {
       if (din == null)
           return;
       din.close();
   }

}