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 Mxn = 4005;
int a[Mxn], b[Mxn];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i], b[i] = a[i];
  sort(b, b + n);
  map<int, int> m;
  int cnt = 0;
  for (int i = 0; i < n; i++) {
    if (m.count(b[i])) continue;
    m[b[i]] = cnt++;
  }
  for (int i = 0; i < n; i++) a[i] = m[a[i]];
  int ans = 1;
  for (int i = 0; i < n; i++) {
    int lst[n], r[n];
    memset(lst, -1, sizeof lst);
    memset(r, 0, sizeof r);
    int cntp = 1, lstp = i;
    for (int j = i + 1; j < n; j++) {
      if (a[j] == a[i]) {
        cntp++, lstp = j;
      } else {
        if (lst[a[j]] == -1)
          r[a[j]] = 2;
        else if (lst[a[j]] < lstp)
          r[a[j]] += 2;
        lst[a[j]] = j;
      }
    }
    int mx = 0;
    for (int i = 0; i < n; i++) {
      if (lst[i] == -1) continue;
      if (lst[i] < lstp)
        mx = max(mx, r[i] + 1);
      else
        mx = max(mx, r[i]);
    }
    ans = max(ans, max(mx, cntp));
  }
  cout << ans << '\n';
  return 0;
}
