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.
/*Author LAVLESH*/
import java.util.*;

public class solution {
	
	public static void main(String args[])
    {
     
     Scanner in=new Scanner(System.in);
     int n=in.nextInt();
     int a[]=new int[4001];
     int dp[][]=new int[4001][4001];
     int ans=0,pv=0;
     for(int i=1;i<=n;i++)a[i]=in.nextInt();
     for(int i=1;i<=n;i++)
     {   pv=0;
    	 for(int j=0;j<i;j++)
     
    	 {
    		dp[j][i]=dp[pv][j]+1;
    		if(a[j]==a[i])pv=j;
    		ans=Math.max(ans,dp[j][i]);
    	 }}
     System.out.println(ans);
  }
 }
