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 maxn = 4005;
int a[maxn];
int dp[maxn][maxn];
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  dp[0][0] = 0;
  int ans = 0;
  for (int i = 1; i <= n; ++i) {
    int l = 0;
    for (int j = 0; j < i; ++j) {
      dp[j][i] = dp[l][j] + 1;
      if (a[i] == a[j]) l = j;
      ans = max(ans, dp[j][i]);
    }
  }
  cout << ans << endl;
  return 0;
}
