Gena loves sequences of numbers. Recently, he has discovered a new type of sequences which he called an almost arithmetical progression. A sequence is an almost arithmetical progression, if its elements can be represented as:

  * a1 = p, where p is some integer; 
  * ai = ai - 1 + ( - 1)i + 1·q (i > 1), where q is some integer. 



Right now Gena has a piece of paper with sequence b, consisting of n integers. Help Gena, find there the longest subsequence of integers that is an almost arithmetical progression.

Sequence s1, s2, ..., sk is a subsequence of sequence b1, b2, ..., bn, if there is such increasing sequence of indexes i1, i2, ..., ik (1 ≤ i1 < i2 < ... < ik ≤ n), that bij = sj. In other words, sequence s can be obtained from b by crossing out some elements.

Input

The first line contains integer n (1 ≤ n ≤ 4000). The next line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106).

Output

Print a single integer — the length of the required longest subsequence.

Examples

Input

2
3 5


Output

2


Input

4
10 20 10 30


Output

3

Note

In the first test the sequence actually is the suitable subsequence. 

In the second test the following subsequence fits: 10, 20, 10.
import java.util.*;
import java.io.*;
import java.util.HashMap;
import java.util.Map;
public class A
{
      public static void main(String ar[]) throws Exception
      {
            BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
            int n=Integer.parseInt(br.readLine());
            String s1[]=br.readLine().split(" ");
            int a[][]=new int[n][2];
            int b[][]=new int[n][2];
            int c[]=new int[n];
            HashMap<Integer,Integer> hm=new HashMap<Integer,Integer>();
            int max=0;
            for(int i=0;i<n;i++)
            {
              a[i][0]=Integer.parseInt(s1[i]);
              a[i][1]=i;
              if(hm.containsKey(a[i][0]))
               hm.put(a[i][0],1+hm.get(a[i][0]));
              else
               hm.put(a[i][0],1);
              max=Math.max(max,hm.get(a[i][0]));
            }
            
            Arrays.sort(a,new Comparator<int[]>(){
                  public int compare(final int f1[],final int f2[])
                  {
                        if(f1[0]>f2[0])
                         return 1;
                        else if(f1[0]==f2[0])
                         return 0;
                        else 
                         return -1;
                  }
            });
            
            int r=-1;
            for(int i=0;i<n;i++)
            {
                  if(i==0 || a[i][0]!=a[i-1][0])
                   r++; 
                  b[i][0]=r; b[i][1]=a[i][1];
            }
            
            Arrays.sort(b,new Comparator<int[]>(){
                  public int compare(final int f1[],final int f2[])
                  {
                        if(f1[1]>f2[1])
                         return 1;
                        else if(f1[1]==f2[1])
                         return 0;
                        else 
                         return -1;
                  }
            });
            
            int dp[][]=new int[n][r+1];
           // int max=1;
            for(int i=0;i<n;i++)
             for(int j=0;j<=r;j++)
              dp[i][j]=1;
            for(int i=0;i<n;i++)
             c[i]=b[i][0];
             
            //System.out.println(Arrays.toString(c));
            for(int i=0;i<n;i++)
            {
                  for(int j=0;j<i;j++)
                  {
                        dp[i][c[j]]=Math.max(dp[i][c[j]],1+dp[j][c[i]]);
                        max=Math.max(max,dp[i][c[j]]);
                  }
            }
            System.out.println(max);
      }
}