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 nmax = 4100;
int dp[nmax][nmax];
int cnt[nmax], n;
int a[nmax], N;
void read() {
  cin >> n;
  N = n;
  for (int j = 0; j < n; ++j) {
    cin >> a[j];
  }
}
void doit(int x) {
  cnt[N - 1] = N;
  for (int i = N - 2; i >= 0; --i) {
    if (a[i + 1] == x)
      cnt[i] = i + 1;
    else
      cnt[i] = cnt[i + 1];
  }
}
void solve() {
  if (n == 1) {
    cout << 1 << endl;
    exit(0);
  }
  int ans = 2;
  for (int i = N - 2; i >= 0; --i) {
    doit(a[i]);
    for (int j = i; j < n; ++j) {
      dp[i][j] = 2;
      if (cnt[j] != N) dp[i][j] = dp[j][cnt[j]] + 1;
      ans = max(ans, dp[i][j]);
    }
  }
  cout << ans << endl;
}
int main() {
  std::ios::sync_with_stdio(false);
  read();
  solve();
  return 0;
}
