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;
vector<int> pos[1000010];
vector<int> b;
int dp[4010][4010];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int temp;
    scanf("%d", &temp);
    b.push_back(temp);
    pos[temp].push_back(i);
  }
  memset(dp, 0, sizeof(dp));
  int ans = 1;
  for (int last = 0; last < n; last++) {
    for (int second_last = 0; second_last < last; second_last++) {
      if (b[last] == b[second_last]) {
        vector<int>::iterator it =
            lower_bound(pos[b[last]].begin(), pos[b[last]].end(), second_last);
        if (it == pos[b[last]].begin()) {
          dp[last][second_last] = 2;
        } else {
          it--;
          dp[last][second_last] = 1 + dp[second_last][*it];
        }
      } else {
        vector<int>::iterator it =
            upper_bound(pos[b[last]].begin(), pos[b[last]].end(), second_last);
        if (it == pos[b[last]].begin()) {
          dp[last][second_last] = 2;
        } else {
          it--;
          dp[last][second_last] = 1 + dp[second_last][*it];
        }
      }
      ans = max(ans, dp[last][second_last]);
    }
  }
  cout << ans;
  return 0;
}
