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.
//package round156;
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.InputMismatchException;

public class A {
	InputStream is;
	PrintWriter out;
	String INPUT = "";

	void solve()
	{
		int n = ni();
		int[] a = na(n);

		int max = 0;
		int[] ind = new int[1000001];
		for(int i = 0;i < n;i++){
			ind[a[i]]++;
			max = Math.max(max, ind[a[i]]);
		}
		int p = 0;
		for(int i = 0;i <= 1000000;i++){
			if(ind[i] > 0){
				ind[i] = p++;
			}
		}
		for(int i = 0;i < n;i++){
			a[i] = ind[a[i]];
		}

		boolean[] ued = new boolean[4000];
		for(int i = 0;i < n;i++){
			if(ued[a[i]])continue;
			ued[a[i]] = true;
			int j = i+1;
			int[] ct = new int[4000];
			while(j < n){
				boolean[] q = new boolean[4000];

				for(;j < n && a[i] != a[j];j++){
					q[a[j]] = true;
				}
				for(int k = 0;k < 4000;k++){
					if(q[k]){
						if(j == n){
							ct[k]++;
						}else{
							ct[k]+=2;
						}
					}
				}
				j++;
			}
			for(int k = 0;k < 4000;k++){
				max = Math.max(max, 1+ct[k]);
			}
		}
		out.println(max);
	}

	void run() throws Exception
	{
		is = oj ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);

		long s = System.currentTimeMillis();
		solve();
		out.flush();
		tr(System.currentTimeMillis()-s+"ms");
	}

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

	private byte[] inbuf = new byte[1024];
	private int lenbuf = 0, ptrbuf = 0;

	private int readByte()
	{
		if(lenbuf == -1)throw new InputMismatchException();
		if(ptrbuf >= lenbuf){
			ptrbuf = 0;
			try { lenbuf = is.read(inbuf); } catch (IOException e) { throw new InputMismatchException(); }
			if(lenbuf <= 0)return -1;
		}
		return inbuf[ptrbuf++];
	}

	private int[] na(int n)
	{
		int[] a = new int[n];
		for(int i = 0;i < n;i++)a[i] = ni();
		return a;
	}

	private int ni()
	{
		int num = 0, b;
		boolean minus = false;
		while((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'));
		if(b == '-'){
			minus = true;
			b = readByte();
		}

		while(true){
			if(b >= '0' && b <= '9'){
				num = num * 10 + (b - '0');
			}else{
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private boolean oj = System.getProperty("ONLINE_JUDGE") != null;
	private void tr(Object... o) { if(!oj)System.out.println(Arrays.deepToString(o)); }
}
