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.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;


public class A {
    
    static StringTokenizer st;
    static BufferedReader in;
    public static void main(String[] args) throws IOException {
        in = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter pw = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
        int n = nextInt();
        int[]a = new int[n+1];
        int[]c = new int[(int) (1e6+1)];
        for (int i = 1; i <= n; i++) {
            a[i] = nextInt();
            c[a[i]]++;
        }
        int ans = 1;
        for (int i = 1; i <= 1e6; i++) {
            ans = Math.max(ans, c[i]);
        }
        int cnt = 0;
        int[]ind = new int[(int) (1e6+1)];
        for (int i = 1; i <= n; i++) {
            if (ind[a[i]]==0) {
                cnt++;
                ind[a[i]] = cnt;
            }
        }
        for (int i = 1; i <= n; i++) {
            a[i] = ind[a[i]];
        }
        int[]leng = new int[n+1];
        int[]last = new int[n+1];
        for (int i = 1; i <= n; i++) {
            Arrays.fill(last, 0);
            int L = i;
            Arrays.fill(leng, 0);
            for (int j = i+1; j <= n; j++) {
                if (a[j]==a[i]) {
                    L = j;
                }
                else {
                    if (last[a[j]] < L) {
                        leng[a[j]] += 2;
                    }
                    last[a[j]] = j;
                }
            }
            boolean[]used = new boolean[n+1];
            for (int j = i+1; j <= n; j++) {
                if (L > last[a[j]] && a[j] != a[i] && !used[a[j]]) {
                    leng[a[j]]++;
                    used[a[j]] = true;
                }
            }
            for (int j = 1; j <= n; j++) {
                if (a[j] != a[i])
                    ans = Math.max(ans, leng[a[j]]);
            }
        }
        System.out.println(ans);
        pw.close();
    }
    private static int nextInt() throws IOException{
        return Integer.parseInt(next());
    }
    
    private static long nextLong() throws IOException{
        return Long.parseLong(next());
    }
    
    private static double nextDouble() throws IOException{
        return Double.parseDouble(next());
    }
    
    private static String next() throws IOException {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
}