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.*;

public class cf256a {
	static int n;
	static int[] v;
	static int[] next;
	static int[][] memo;
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		n = in.nextInt();
		v = new int[n];
		for(int i=0; i<n; i++) v[i] = in.nextInt();
		next = new int[n];
		Arrays.fill(next,n);
		for(int i=0; i<n; i++)
			for(int j=n-1; j>i; j--)
				if(v[i]==v[j])
					next[i]=j;
		memo = new int[n+1][n+1];
		for(int[] x : memo) Arrays.fill(x,-1);
		int ans = 1;
		for(int i=n-1; i>=0; i--)
			for(int j=n-1; j>i; j--)
				ans = Math.max(ans,go(i,j));
		System.out.println(ans);
	}
	static int go(int i, int j) {
		if(i > j) return 1 + go(j,i);
		if(j == n) return 1;
		if(memo[i][j] != -1) return memo[i][j];
		return memo[i][j] = go(next[i],j);
	}
}
