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 main() {
  long long int t, n, m, f, a, b, c, l, q, k, x, y, z, ans, mn, mx, sum = 0, p;
  cin >> n;
  map<long long int, vector<long long int> > mp;
  vector<long long int> t1, t2;
  vector<long long int>::iterator it;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &t);
    mp[t].push_back(i);
  }
  ans = 0;
  for (auto it1 = mp.begin(); it1 != mp.end(); it1++) {
    for (auto it2 = mp.begin(); it2 != mp.end(); it2++) {
      if (it1->first == it2->first) {
        t1 = it1->second;
        ans = max(ans, (long long int)t1.size());
        continue;
      }
      t1 = it1->second;
      t2 = it2->second;
      f = 1;
      sum = 1;
      x = 0;
      y = 0;
      while (1) {
        z = sum;
        if (f) {
          while (y < t2.size()) {
            if (t2[y] > t1[x]) {
              sum++;
              x++;
              break;
            }
            y++;
          }
        } else {
          while (x < t1.size()) {
            if (t1[x] > t2[y]) {
              sum++;
              y++;
              break;
            }
            x++;
          }
        }
        if (z < sum) {
          f = 1 - f;
        } else {
          break;
        }
      }
      ans = max(ans, sum);
    }
  }
  cout << ans << endl;
}
