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 n, r = 0, ans = 1, a[4007], freq[1000007], dp[4007][4007];
unordered_map<int, int> um;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  std::cin >> n;
  for (int i = 0; i < n; i++) std::cin >> a[i], freq[a[i]]++;
  for (int i = 0; i < 1000007; i++)
    if (freq[i]) r++, freq[i] = r;
  for (int i = 0; i < n; i++) a[i] = freq[a[i]] - 1;
  for (int i = 0; i < n; i++) {
    int k = -1;
    for (int j = 0; j < i; j++) {
      if (k == -1)
        dp[i][j] = 2;
      else
        dp[i][j] = dp[j][k] + 1;
      if (a[i] == a[j]) k = j;
      ans = (((ans) > (dp[i][j])) ? (ans) : (dp[i][j]));
    }
  }
  printf("%d\n", ans);
}
