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 int N = 4e3 + 3;
int dp[N][N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout << setprecision(25);
  int t = 1;
  while (t--) {
    int n;
    cin >> n;
    vector<int> b(n);
    for (int i = 0; i < n; i++) {
      cin >> b[i];
    }
    int ind = 0;
    map<int, int> m;
    for (int i = 0; i < n; i++) {
      if (m.find(b[i]) == m.end()) m[b[i]] = ind++;
    }
    for (int i = 0; i < n; i++) b[i] = m[b[i]];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) {
        dp[i][j] = 1;
      }
    }
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < i; j++) {
        dp[i][b[j]] = max(1 + dp[j][b[i]], dp[i][b[j]]);
      }
    }
    int ans = 0;
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < n; j++) ans = max(ans, dp[i][j]);
    }
    cout << ans << "\n";
  }
}
