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;
template <typename T>
int compressed(vector<T> &a) {
  vector<T> b = a;
  sort(begin(b), end(b));
  b.erase(unique(begin(b), end(b)), end(b));
  for (T &ai : a) ai = lower_bound(begin(b), end(b), ai) - begin(b);
  return b.size();
}
int main() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (auto &ai : a) cin >> ai;
  int sz = compressed(a);
  vector<vector<int>> dp(sz, vector<int>(sz));
  for (int ai : a)
    for (int x = 0; x < sz; x++) dp[x][ai] = dp[ai][x] + 1;
  int ans = 0;
  for (auto x : dp) ans = max(ans, *max_element(begin(x), end(x)));
  cout << ans << endl;
  return 0;
}
