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 int dp[4001][4001];
void pre() {}
void solve() {
  int n;
  cin >> n;
  vector<long long int> a(n);
  set<long long int> s;
  for (int i = 0; i < (int)(n); ++i) {
    cin >> a[i];
    s.insert(a[i]);
  }
  int cnt = 1;
  map<long long int, int> make_pair;
  while (!s.empty()) {
    make_pair[*s.begin()] = cnt++;
    s.erase(*s.begin());
  }
  for (int i = 0; i < (int)(n); ++i) a[i] = make_pair[a[i]];
  for (int i = 0; i < (int)(n); ++i) {
    dp[a[i]][a[i]]++;
    set<int> st;
    for (int j = 0; j < (int)(i); ++j) {
      if (a[j] != a[i]) {
        st.insert(a[j]);
      }
    }
    for (int j : st) {
      if (dp[j][a[i]] % 2 == 1) {
        dp[j][a[i]]++;
      }
      if (dp[a[i]][j] % 2 == 0) {
        dp[a[i]][j]++;
      }
    }
    for (int j = 1; j <= (int)(cnt - 1); ++j)
      if (dp[a[i]][j] == 0) dp[a[i]][j] = 1;
  }
  long long int ans = -1;
  for (int i = 1; i <= (int)(cnt - 1); ++i) {
    for (int j = 1; j <= (int)(cnt - 1); ++j) {
      ans = max(ans, dp[i][j]);
    }
  }
  cout << ans;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cout.precision(10);
  cout << fixed;
  pre();
  int T = 1;
  while (T--) {
    solve();
  }
  return 0;
}
