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 Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
long long hell = round(1e18);
signed solve() {
  long long n;
  cin >> n;
  map<long long, long long> m;
  vector<long long> v(n);
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
    if (m[v[i]] == 0) m[v[i]] = i;
  }
  vector<vector<long long> > dp(n, vector<long long>(n, 1));
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < i; j++) {
      dp[i][m[v[j]]] = max(1 + dp[j][m[v[i]]], dp[i][m[v[j]]]);
    }
  }
  long long mx = 0;
  for (long long i = 0; i < n; i++)
    for (long long j = 0; j < n; j++) mx = max(dp[i][j], mx);
  cout << mx;
  return 0;
}
signed main() {
  long long t = 1;
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  while (t--) {
    solve();
  }
  return 0;
}
