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;
void solve() {
  long long n, i, j, k, x, y, z;
  cin >> n;
  vector<long long> a, temp;
  for (i = 0; i < n; i++) {
    cin >> x;
    a.push_back(x);
    temp.push_back(x);
  }
  map<int, int> m;
  long long ind = 0;
  for (i = 0; i < n; i++) {
    x = a[i];
    if (m.find(a[i]) == m.end()) {
      m[a[i]] = ind++;
    }
  }
  for (i = 0; i < n; i++) {
    a[i] = m[a[i]];
  }
  vector<vector<int> > dp;
  long long si = 4005;
  vector<int> c(si);
  dp.resize(si, c);
  for (i = 0; i < si; i++)
    for (j = 0; j < si; j++) dp[i][j] = 1;
  for (i = 0; i < n; i++) {
    for (j = 0; j < i; j++) {
      dp[i][a[j]] = max(dp[i][a[j]], 1 + dp[j][a[i]]);
    }
  }
  long long ans = -1;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      ans = max(dp[i][j], (int)ans);
    }
  }
  cout << ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  while (t-- > 0) {
    solve();
  }
}
