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

public class Codeforces
{
    public static void main(String args[])throws Exception
    {
        BufferedReader bu=new BufferedReader(new InputStreamReader(System.in));
        StringBuilder sb=new StringBuilder();
        int n=Integer.parseInt(bu.readLine());
        String s[]=bu.readLine().split(" ");
        int i,a[]=new int[n],j; long sum[]=new long[n];
        HashMap<Integer,Integer> hm=new HashMap<>();
        j=0;
        for(i=0;i<n;i++)
        {
            a[i]=Integer.parseInt(s[i]);
            if(hm.get(a[i])==null) hm.put(a[i],j++);
        }
        for(i=0;i<n;i++)
        a[i]=hm.get(a[i]);

        int dp[][]=new int[n][n];
        for(i=0;i<n;i++)
        for(j=0;j<n;j++)
        dp[i][j]=1;

        for(i=0;i<n;i++)
        for(j=0;j<i;j++)
        dp[i][a[j]]=Math.max(1+dp[j][a[i]],dp[i][a[j]]);
        int ans=0;
        for(i=0;i<n;i++)
        for(j=0;j<n;j++) ans=Math.max(ans,dp[i][j]);
        System.out.print(ans);
    }
}