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 f[4001][4001];
int main() {
  int n, i, j, s, ans = 1;
  int a[4001], b[4001];
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> a[i];
    b[i] = a[i];
  }
  sort(b, b + n);
  map<int, int> m;
  map<int, int>::iterator itr;
  s = unique(b, b + n) - b;
  for (i = 1; i <= s; i++) m.insert(make_pair(b[i - 1], i));
  for (i = 0; i < n; i++) {
    a[i] = m[a[i]];
  }
  for (i = 1; i < n; i++) {
    for (j = 0; j < i; j++) {
      f[i][a[j]] = max(max(f[i][a[j]], f[j][a[i]] + 1), 2);
      ans = max(ans, f[i][a[j]]);
    }
  }
  cout << ans;
  return 0;
}
