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 double pi = acos(-1);
const int n_ = 1e6 + 5;
int n, arr[4005], ans = 1, vis[4005][4005], id;
vector<int> p[4005];
set<int> s;
map<int, int> m;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &arr[i]);
    s.insert(arr[i]);
  }
  for (int i : s) m[i] = id++;
  for (int i = 0; i < n; i++) {
    arr[i] = m[arr[i]];
    p[arr[i]].push_back(i);
  }
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      if (vis[arr[i]][arr[j]]) continue;
      vis[arr[i]][arr[j]] = vis[arr[j]][arr[i]] = 1;
      if (arr[i] == arr[j]) {
        ans = max(ans, (int)p[arr[i]].size());
        continue;
      }
      int a = 0, b = -1, f = 1, cnt = 1;
      while (a < p[arr[i]].size() && b < (int)p[arr[j]].size()) {
        if (f) {
          ++b;
          while (b < p[arr[j]].size() && p[arr[j]][b] < p[arr[i]][a]) ++b;
          if (b == p[arr[j]].size()) break;
        } else {
          ++a;
          while (a < p[arr[i]].size() && p[arr[i]][a] < p[arr[j]][b]) ++a;
          if (a == p[arr[i]].size()) break;
        }
        ++cnt;
        f = !f;
      }
      ans = max(ans, cnt);
    }
  }
  printf("%d", ans);
  return 0;
}
