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;
vector<vector<long long int>> cn(long long int n, long long int m) {
  return vector<vector<long long int>>(n, vector<long long int>(m));
}
bool compare(char &s1, char &s2) { return s1 > s2; }
bool sortmahstyle(const pair<long long int, long long int> &a,
                  const pair<long long int, long long int> &b) {
  if (a.first < b.first) return true;
  if (a.first == b.first && a.second != b.second) return true;
  return false;
}
long long int const mod = 998244353;
long long int const inf = 1e18;
long long int const maxn = 1e6 + 1;
void solve() {
  long long int n;
  cin >> n;
  long long int a[n + 1];
  for (long long int i = 0; i < n; i++) cin >> a[i + 1];
  long long int it = 1;
  long long int id[maxn] = {0};
  for (long long int i = 1; i < n + 1; i++)
    if (id[a[i]] == 0) id[a[i]] = it++;
  long long int dp[n + 1][n + 1];
  for (long long int i = 1; i < n + 1; i++)
    for (long long int j = 1; j < n + 1; j++) dp[i][j] = 1;
  for (long long int i = 1; i <= n; i++) {
    for (long long int j = 1; j < i; j++) {
      dp[i][id[a[j]]] = max(dp[i][id[a[j]]], 1 + dp[j][id[a[i]]]);
    }
  }
  long long int ans = 0;
  for (long long int i = 1; i <= n; i++)
    for (long long int j = 1; j <= n; j++) ans = max(ans, dp[i][j]);
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  t = 1;
  for (long long int testcases = 0; testcases < t; testcases++) {
    solve();
  }
  return 0;
}
