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;
long long int dp[4003][4003], m[1000003], cnt[1000006];
int main() {
  long long int n, b[4003], i, j, k;
  long long int ans = 0;
  cin >> n;
  for (i = 1; i <= n; i++) {
    cin >> b[i];
    cnt[b[i]]++;
    ans = max(ans, cnt[b[i]]);
  }
  for (i = 1; i <= n; i++) {
    long long int prev = 0;
    for (j = 0; j < i; j++) {
      dp[i][j] = dp[j][prev] + 1;
      if (b[i] == b[j]) prev = j;
    }
  }
  for (i = 0; i <= n; i++)
    for (j = 0; j <= n; j++) ans = max(ans, dp[i][j]);
  cout << ans << endl;
}
