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(false);
  cin.tie(NULL);
  int n, v;
  cin >> n;
  unordered_map<int, int> mapIdx;
  vector<int> vals;
  int idx = 0;
  for (int i = 0; i < n; i++) {
    cin >> v;
    if (!mapIdx.count(v)) mapIdx[v] = idx++;
    vals.push_back(mapIdx[v]);
  }
  vector<vector<int>> dp(n, vector<int>(idx, 1));
  int res = 1;
  for (int i = 0; i < n; i++) {
    v = vals[i];
    for (int j = 0; j < i; j++) {
      dp[i][vals[j]] = max(dp[i][vals[j]], dp[j][v] + 1);
      res = max(dp[i][vals[j]], res);
    }
  }
  cout << res << "\n";
}
