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 eval(vector<int> &a, vector<int> &b) {
  int n = a.size(), m = b.size(), ret = 0, idx = -1, which = 0;
  int i = 0, j = 0;
  while (1) {
    if (which == 0) {
      while (1) {
        if (i >= n) return ret;
        if (a[i] > idx) {
          idx = a[i];
          i++;
          ret++;
          which = (which + 1) % 2;
          break;
        }
        i++;
      }
    } else {
      while (1) {
        if (j >= m) return ret;
        if (b[j] > idx) {
          idx = b[j];
          j++;
          ret++;
          which = (which + 1) % 2;
          break;
        }
        j++;
      }
    }
  }
  return ret;
}
const int mx = 1e6 + 5;
map<int, vector<int> > mp;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> v(n);
  int ret = 0;
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    v[i] = x;
    mp[x].push_back(i);
    ret = max(ret, (int)mp[x].size());
  }
  vector<pair<int, int> > w;
  for (map<int, vector<int> >::iterator it = mp.begin(); it != mp.end(); it++)
    w.push_back(make_pair((*it).second.size(), (*it).first));
  sort(w.begin(), w.end(), greater<pair<int, int> >());
  int sz = w.size();
  vector<int> cur;
  for (int i = 0; i < sz; i++) cur.push_back(w[i].second);
  int m = cur.size();
  for (int i = 0; i < m; i++) {
    for (int j = i + 1; j < m; j++) {
      vector<int> a = mp[cur[i]], b = mp[cur[j]];
      int x = a.size(), y = b.size();
      if (ret >= x + y) break;
      if (a[0] > b[0]) swap(a, b);
      ret = max(ret, eval(a, b));
    }
  }
  cout << ret << endl;
  return 0;
}
