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 MX = 1e6 + 100, Mx = 4 * 1e3 + 100;
int a[Mx], dp[Mx][Mx], an[MX], ann[MX], n;
void input() {
  cin >> n;
  int num = 1;
  for (int i = 1; i <= n; ++i) {
    int e;
    cin >> e;
    if (ann[e] == 0) {
      a[i] = num;
      ann[e] = num;
      an[e]++;
      num++;
    } else {
      a[i] = ann[e];
      an[e]++;
    }
  }
}
void solve() {
  for (int i = 1; i <= n; ++i) {
    dp[a[i]][0] = 1;
    for (int j = i - 1; j > 0; --j) {
      if (a[i] == a[j]) continue;
      dp[a[i]][a[j]] = max(dp[a[j]][a[i]] + 1, 2);
    }
  }
  int ans = 1;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) ans = max(ans, dp[i][j]);
  }
  for (int i = 1; i < MX; ++i) ans = max(ans, an[i]);
  cout << ans << endl;
}
int main() {
  input();
  solve();
}
