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.*;
public class R156D2C {
 
	public static void main(String[] args) throws Exception{
		// TODO Auto-generated method stub
 		BufferedReader bf = new BufferedReader(new InputStreamReader(System.in));
 		PrintWriter out = new PrintWriter(System.out);
 		int n = Integer.parseInt(bf.readLine());
 		Set<Integer> tree = new TreeSet<Integer>();
 		int[] array = new int[n];
 		Map<Integer, Integer> amap = new HashMap<Integer, Integer>();
 		StringTokenizer st = new StringTokenizer(bf.readLine());
 		for(int i = 0;i<n;i++){
 			array[i] = Integer.parseInt(st.nextToken());
 			tree.add(array[i]);
 		}
 		int count = 0;
 		for (int i : tree){
 			amap.put(i, count);
 			count+=1;
 		}
 		for(int i = 0;i<n;i++){
 			array[i] = amap.get(array[i]);
 		}
 		
 		int[][] dp = new int[n][n];
 		for(int i = 0;i<=n-1;i++){
 			for(int j = 0;j<=n-1;j++){
 				dp[i][j] = 1;
 			}
 		}
 		
 		for(int i = 1;i<=n-1;i++){
 			for(int j = 0;j<i;j++){
 				dp[i][array[j]] = Math.max(1+dp[j][array[i]], dp[i][array[j]]);
 			}
 		}
 		
 		int max = Integer.MIN_VALUE;
 		for(int i = 0;i<=n-1;i++){
 			for(int j = 0;j<=n-1;j++){
 				if (dp[i][j] > max)
 					max = dp[i][j];
 			}
 		}
 		out.println(max);
 		out.close();
 		
 		
 		
	}
 
}