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;
long long A[4000];
vector<int> O[1000001];
pair<long long, long long> longest[1000001];
set<int> val;
int main() {
  ios::sync_with_stdio(false);
  int N;
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
    O[A[i]].push_back(i);
    val.insert(A[i]);
  }
  if (val.size() == 1) {
    cout << 1;
    return 0;
  }
  if (val.size() == N) {
    cout << 2;
    return 0;
  }
  long long best = 0;
  vector<int> P(val.begin(), val.end());
  for (int i = 0; i < P.size(); i++) {
    for (int j = 0; j < P.size(); j++) longest[P[j]] = make_pair(0, 0);
    int u = P[i];
    long long umax = O[u].size();
    int j = O[u][0];
    int k = 0;
    best = max(best, umax);
    for (; j < N; j++) {
      if (A[j] == u) {
        k++;
        continue;
      }
      pair<long long, long long> longestj = longest[A[j]];
      if (k - longestj.second == 0) {
        continue;
      } else {
        longest[A[j]] = make_pair(longestj.first + 1, k);
        best = max(best, 2 * longest[A[j]].first +
                             ((longest[A[j]].second == umax) ? 0 : 1));
      }
    }
  }
  cout << best;
  return 0;
}
