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.*;
public class Main{
//public class math{
	public static void main(String[] args)throws IOException{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		int n=Integer.parseInt(br.readLine());
		String s=br.readLine();
		String str[]=s.split(" ");
		int arr[]=new int[n+1];
		for(int i=1;i<=n;i++){
			arr[i]=Integer.parseInt(str[i-1]);
		}
		
		int dp[][]=new int[n+1][n+1];
		
		int ans=0;
		for(int i=1;i<=n;i++){
			int k=0;
			for(int j=0;j<i;j++){
				dp[j][i]=dp[k][j]+1;
				if(arr[i]==arr[j]) k=j;
				ans=Math.max(ans,dp[j][i]);
			}
//			System.out.println("DSF");
		}
/*		for(int i=1;i<=n;i++){
			for(int j=1;j<=n;j++){
				System.out.print(dp[i][j]+" ");	
			}	
			System.out.println();
		}
*/
		System.out.println(ans);

	}	
}
