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.
#include <bits/stdc++.h>
using namespace std;
int n, a[10100], v[1001000];
int p[10100], nx[10100], ans = 1;
int solve(int x, int y, int z) {
  int t = x;
  while (t <= y) t = nx[t];
  if (t <= n) return solve(y, t, z + 1);
  return z;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    p[i] = v[a[i]];
    v[a[i]] = i;
  }
  for (int i = 1; i <= n; i++) {
    nx[i] = n + 1;
    for (int j = i + 1; j <= n; j++)
      if (a[j] == a[i]) {
        nx[i] = j;
        break;
      }
  }
  for (int i = 1; i <= n; i++)
    if (!p[i])
      for (int j = i + 1; j <= n; j++) {
        if (p[j] > i) continue;
        ans = max(ans, solve(i, j, 2));
      }
  cout << ans << endl;
}
