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 N = 4e3 + 5;
const long long A = 1e6 + 5;
long long a[N], len[N][N], ord[A];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, ai, aj, ans;
  cin >> n;
  for (long long i = 0; i < n; i++) cin >> a[i];
  for (long long i = 0; i < n; i++) ord[a[i]] = 1;
  for (long long i = 1; i < A; i++) ord[i] += ord[i - 1];
  for (long long i = 0; i < n; i++)
    for (long long j = 1; j <= ord[A - 1]; j++) {
      ai = ord[a[i]];
      aj = j;
      len[aj][ai] = 1 + len[ai][aj];
    }
  ans = 0;
  for (long long i = 1; i <= ord[A - 1]; i++)
    for (long long j = 1; j <= ord[A - 1]; j++) ans = max(ans, len[i][j]);
  cout << ans << "\n";
  return 0;
}
