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;
const int MAXN = 1e6 + 9;
vector<int> idx[MAXN];
vector<bool> vis(MAXN);
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    idx[a[i]].push_back(i);
  }
  int res = 1;
  for (int i = 0; i < n; i++) {
    if (vis[a[i]]) continue;
    vis[a[i]] = true;
    for (int j = i + 1; j < n; j++) {
      int ind = i, cur = 1;
      for (int k = 1;; k++) {
        if (k & 1) {
          auto it = upper_bound(idx[a[j]].begin(), idx[a[j]].end(), ind);
          if (it == idx[a[j]].end()) break;
          ind = *it;
          ++cur;
        } else {
          auto it = upper_bound(idx[a[i]].begin(), idx[a[i]].end(), ind);
          if (it == idx[a[i]].end()) break;
          ind = *it;
          ++cur;
        }
      }
      res = max(res, cur);
    }
  }
  cout << res << '\n';
  return 0;
}
