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 long long int MOD = 1000000007;
const long double PI = 2 * acos(0.0);
const vector<long long int> dx = {1, -1, 0, 0};
const vector<long long int> dy = {0, 0, 1, -1};
vector<long long int> ga(long long int n, bool oneIndexed = false) {
  vector<long long int> a;
  if (oneIndexed) a.push_back(0ll);
  for (long long int i = 0; i < n; i++) {
    long long int p;
    cin >> p;
    a.push_back(p);
  }
  return move(a);
}
vector<string> gas(unsigned long long int n, bool oneIndexed = false) {
  vector<string> a;
  if (oneIndexed) a.push_back("");
  for (unsigned long long int i = 0; i < n; i++) {
    string p;
    cin >> p;
    a.push_back(p);
  }
  return move(a);
}
template <typename T, typename A>
void pa(vector<T, A> const &a, long long int begin = 0,
        long long int end = -1) {
  if (end == -1) end = (long long int)a.size() - 1;
  for (long long int i = begin; i <= end; i++) {
    cout << a[i] << " ";
  }
  cout << "\n";
}
template <typename T, typename A>
void papair(vector<T, A> const &a, long long int begin = 0,
            long long int end = -1) {
  if (end == -1) end = (long long int)a.size() - 1;
  for (long long int i = begin; i <= end; i++) {
    cout << a[i].first << " " << a[i].second << "\n";
  }
}
vector<vector<long long int> > adjList(1e6 + 1, vector<long long int>(0));
set<long long int> dist;
void solve() {
  long long int n;
  cin >> n;
  vector<long long int> a = ga(n);
  for (long long int i = 0; i < n; i++) {
    adjList[a[i]].push_back(i);
    dist.insert(a[i]);
  }
  long long int ans = 0;
  for (auto a : dist) {
    for (auto b : dist) {
      auto curr = adjList[a][0];
      long long int check = 1, len = 1;
      while (true) {
        if (check) {
          auto i2 = lower_bound(adjList[b].begin(), adjList[b].end(), curr + 1);
          if (i2 == adjList[b].end()) break;
          curr = *i2;
        } else {
          auto i2 = lower_bound(adjList[a].begin(), adjList[a].end(), curr + 1);
          if (i2 == adjList[a].end()) break;
          curr = *i2;
        }
        check = !check;
        len++;
      }
      if (len > ans) ans = len;
      ;
    }
  }
  cout << ans;
}
int main(void) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}
