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;
map<int, vector<int> > m;
int count(vector<int> &v1, vector<int> &v2) {
  int temp = v1[0], counter = 1, i = 1, j = 0, b = 0;
  while (i < v1.size() || j < v2.size()) {
    if (!(!b && j < v2.size()) && !(b && i < v1.size())) break;
    while (!b && j < v2.size()) {
      if (v2[j] > temp) {
        temp = v2[j];
        counter++;
        b++;
        break;
      }
      j++;
    }
    while (b && i < v1.size()) {
      if (v1[i] > temp) {
        temp = v1[i];
        counter++;
        b--;
        break;
      }
      i++;
    }
  }
  return counter;
}
int main() {
  int n, temp;
  int j1 = 1, j2 = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> temp;
    m[temp].push_back(j1);
    j1++;
  }
  int res = 1;
  for (map<int, vector<int> >::iterator it = m.begin(); it != m.end(); it++)
    for (map<int, vector<int> >::iterator jt(it); jt != m.end(); jt++) {
      res = max(res, count(jt->second, it->second));
      res = max(res, count(it->second, jt->second));
    }
  cout << res << endl;
  return 0;
}
