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;
int n, dp[4000][4000][2], v[4000], last[1000001], need, res = 1;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", v + i);
  for (int i = n - 1; i >= 0; --i) {
    for (int j = n - 1; j >= i; --j) last[v[j]] = -1;
    for (int j = n - 1; j > i; --j) {
      for (int o = 0; o < 2; ++o) {
        need = o ? v[i] - v[j] : v[j] - v[i];
        need *= o ? 1 : -1;
        need += v[j];
        if (need >= 1 && need <= 1000000 && last[need] != -1)
          dp[i][j][o] = 1 + dp[j][last[need]][!o];
        if (o && res < dp[i][j][o] + 2) res = dp[i][j][o] + 2;
      }
      last[v[j]] = j;
    }
  }
  printf("%d\n", res);
  return 0;
}
