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>
int a[4010], b[4010], c[4010][4010], f[4010][4010], i, j, n, m, ans;
int main() {
  scanf("%d\n", &n);
  for (i = 1; i <= n; ++i) scanf("%d", &a[i]), b[i] = a[i];
  std::sort(b + 1, b + n + 1);
  m = std::unique(b + 1, b + n + 1) - b - 1;
  for (i = 1; i <= n; ++i) {
    a[i] = std::lower_bound(b + 1, b + m + 1, a[i]) - b;
    for (j = 1; j <= m; ++j) c[i][j] = c[i - 1][j];
    c[i][a[i]] = i;
  }
  ans = 1;
  for (i = 1; i <= n; ++i)
    for (f[i][0] = 1, j = 1; j < i; ans = std::max(ans, f[i][j++]))
      f[i][j] = f[j][c[j - 1][a[i]]] + 1;
  printf("%d\n", ans);
}
