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;
int main() {
  ios_base::sync_with_stdio(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  vector<int> r;
  for (int i = 0; i < n; ++i) {
    bool jest = false;
    for (int j = 0; j < i; ++j)
      if (a[i] == a[j]) jest = true;
    if (!jest) r.push_back(a[i]);
  }
  vector<int> gdzie(1000001, -1);
  for (int i = 0; i < r.size(); ++i) gdzie[r[i]] = i;
  int odp = n == 1 ? 1 : 2;
  for (int i = 0; i < n; ++i) {
    vector<vector<int> > przedzialy(r.size());
    vector<int> nry;
    for (int j = i; j < n; ++j)
      if (a[i] == a[j]) nry.push_back(j);
    if (nry.size() == 1) continue;
    nry.push_back(n);
    for (int j = 0; j < (int)nry.size() - 1; ++j) {
      for (int k = nry[j] + 1; k < nry[j + 1]; ++k) {
        int nr = gdzie[a[k]];
        if (przedzialy[nr].size() == 0 || przedzialy[nr].back() != nry[j])
          przedzialy[nr].push_back(nry[j]);
      }
    }
    nry.pop_back();
    for (int j = 0; j < przedzialy.size(); ++j) {
      if (przedzialy[j].size() != 0) {
        if (przedzialy[j].back() == nry.back())
          odp = max(odp, (int)przedzialy[j].size() * 2);
        else
          odp = max(odp, (int)przedzialy[j].size() * 2 + 1);
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    int il = 0;
    for (int j = 0; j < n; ++j)
      if (a[i] == a[j]) ++il;
    odp = max(odp, il);
  }
  cout << odp;
  return 0;
}
