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 MOD = 998244353;
const long long INF = 1e18;
const long double PI = 4 * atan((long double)1);
const int INFTY = 1e7;
int N;
long long B[4001];
set<int> second;
unordered_map<int, int> m;
long long dp[4001][4001];
int main() {
  cin >> N;
  int index = 0;
  for (int i = 0; i < (N); i++) cin >> B[i], second.insert(B[i]);
  for (auto v : second) m[v] = index++;
  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;
  long long best = 1;
  for (int i = 0; i < (N); i++) {
    for (int j = 0; j < i; j++) {
      dp[i][B[j]] = ((1 + dp[j][B[i]]) > (dp[i][B[j]]) ? (1 + dp[j][B[i]])
                                                       : (dp[i][B[j]]));
      best = ((dp[i][B[j]]) > (best) ? (dp[i][B[j]]) : (best));
    }
  }
  cout << best << "\n";
}
