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 main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  int a[n + 1];
  int mn = INT_MIN;
  map<int, int> m;
  for (int i = 1; i <= n; i++) cin >> a[i], m[a[i]]++;
  int cnt = 1;
  for (auto &i : m) i.second = cnt, cnt++;
  for (int i = 1; i <= n; i++) a[i] = m[a[i]];
  unordered_map<int, vector<int>> mp;
  for (int i = 1; i <= n; i++) mp[a[i]].push_back(i);
  int ans = 1;
  for (auto &i : mp) {
    for (auto &j : mp) {
      if (i.first == j.first)
        ans = max(ans, (int)(i.second).size());
      else {
        int cnt = 0;
        int p = 0, q = 0, last1 = 0, last2 = 0;
        while (1) {
          while (p < i.second.size() && i.second[p] <= last2) p++;
          if (p == i.second.size())
            break;
          else {
            cnt++;
            last1 = i.second[p];
            p++;
          }
          while (q < j.second.size() && j.second[q] <= last1) q++;
          if (q == (j.second).size())
            break;
          else {
            cnt++;
            last2 = j.second[q];
            q++;
          }
        }
        ans = max(ans, cnt);
      }
    }
  }
  cout << ans;
}
