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 Main {
	boolean eof;

	String nextToken() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (Exception e) {
				eof = true;
				return "-1";
			}
		}
		return st.nextToken();
	}

	int nextInt() {
		return Integer.parseInt(nextToken());
	}

	long nextLong() {
		return Long.parseLong(nextToken());
	}

	double nextDouble() {
		return Double.parseDouble(nextToken());
	}

	void solve() {
		int n = nextInt();
		int[] a = new int[n];
		for (int i = 0; i < n; ++i) {
			a[i] = nextInt() - 1;
		}
		int[] b = new int[1000000];
		Arrays.fill(b, -1);
		ArrayList<Integer> l = new ArrayList<Integer>();
		
		for (int i = 0; i < n; ++i) {
			if (b[a[i]] == -1) {
				l.add(a[i]);
				b[a[i]] = l.size() - 1;
		
			}
		}
		int kol = l.size();
		int[][] d = new int[kol][kol];
		int ans = 0;
		boolean[] was = new boolean[kol];

		for (int i = 0; i < n; ++i) {
			Arrays.fill(was, false);
			for (int j = 0; j < i; ++j) {
				if (!was[b[a[j]]]) {
					d[b[a[i]]][b[a[j]]] = Math.max(d[b[a[i]]][b[a[j]]],
							d[b[a[j]]][b[a[i]]] + 1);
				}
				was[b[a[j]]] = true;
				ans = Math.max(ans, d[b[a[i]]][b[a[j]]]);
			}

		}
		out.print(ans + 1);
	}

	BufferedReader br;
	StringTokenizer st;
	PrintWriter out;

	void run() {
		try {
			br = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(new OutputStreamWriter(System.out));

			// br = new BufferedReader(new FileReader("input.txt"));
			// out = new PrintWriter(new FileWriter("output.txt"));

			solve();

			br.close();
			out.close();
		} catch (Throwable e) {
			e.printStackTrace();
			System.exit(1);
		}
	}

	public static void main(String[] args) {
		new Main().run();
	}
}