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 N = 4005, mod = 998244353;
int n, a[N], res;
vector<int> G[1000005], dis;
set<int> S;
int main() {
  cin.tie(0);
  cout.tie(0);
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) G[a[i]].push_back(i);
  for (int i = 0; i < n; i++)
    if (S.count(a[i]) == 0) {
      S.insert(a[i]);
      dis.push_back(a[i]);
    }
  for (int i = 0; i < ((int)dis.size()); i++)
    for (int j = 0; j < ((int)dis.size()); j++) {
      int r, s, k = 0, c = 0;
      bool t = true;
      r = s = 0;
      while (r < ((int)G[dis[i]].size()) || s < ((int)G[dis[j]].size())) {
        if (t) {
          while (r < ((int)G[dis[i]].size()) && G[dis[i]][r] < k) r++;
          if (r < ((int)G[dis[i]].size())) {
            k = G[dis[i]][r] + 1;
            c++;
            t = false;
          } else {
            r = mod;
            s = mod;
          }
        } else {
          while (s < ((int)G[dis[j]].size()) && G[dis[j]][s] < k) s++;
          if (s < ((int)G[dis[j]].size())) {
            k = G[dis[j]][s] + 1;
            c++;
            t = true;
          } else {
            r = s = mod;
          }
        }
      }
      res = max(res, c);
    }
  cout << res;
  return 0;
}
