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;
long long a[5000], inf = 9e18;
vector<set<long long> > ind(1000005);
set<long long>::iterator ii;
set<long long> t;
vector<long long> uniq;
int main() {
  long long n, m, i, j;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> a[i];
    ind[a[i]].insert(i);
  }
  long long ans = 0;
  for (i = 1; i <= 1000000; i++) {
    ind[i].insert(inf);
    if (ind[i].size() > 1) {
      t.insert(i);
      ans = max(ans, (long long)ind[i].size() - 1);
    }
  }
  for (ii = t.begin(); ii != t.end(); ii++) uniq.push_back((*ii));
  for (i = 0; i < uniq.size(); i++) {
    for (j = 0; j < uniq.size(); j++) {
      long long one = uniq[i], two = uniq[j];
      if (two != one) {
        long long cur = 0, lastpos = 0;
        while (true) {
          if (cur % 2 == 1)
            ii = ind[two].upper_bound(lastpos);
          else
            ii = ind[one].upper_bound(lastpos);
          if ((*ii) == inf) break;
          cur++;
          lastpos = (*ii);
        }
        ans = max(ans, cur);
      }
    }
  }
  cout << ans;
  return 0;
}
