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 INF = 0x3f3f3f3f;
const int MOD7 = (int)1e9 + 7;
template <class T1, class T2>
int upmin(T1 &x, T2 v) {
  if (x > v) {
    x = v;
    return 1;
  }
  return 0;
}
template <class T1, class T2>
int upmax(T1 &x, T2 v) {
  if (x < v) {
    x = v;
    return 1;
  }
  return 0;
}
int N;
vector<int> B;
void init() {
  cin >> N;
  B = vector<int>(N);
  for (int i = 0; i < N; ++i) cin >> B[i];
}
map<int, vector<int> > v_pos;
vector<vector<int> > dp;
void preprocess() {
  for (int i = 0; i < N; ++i) v_pos[B[i]].emplace_back(i);
  dp = vector<vector<int> >(N, vector<int>(N, 2));
  for (int i = 0; i < N; ++i) dp[i][i] = 1;
  for (int i = 0; i < N; ++i)
    for (int j = i + 1; j < N; ++j) {
      auto it = upper_bound(v_pos[B[i]].begin(), v_pos[B[i]].end(), j);
      if (it != v_pos[B[i]].end()) upmax(dp[j][*it], dp[i][j] + 1);
    }
}
void solve() {
  int ans = 1;
  for (int i = 0; i < N; ++i)
    for (int j = i + 1; j < N; ++j) upmax(ans, dp[i][j]);
  cout << ans << endl;
}
signed main() {
  ios::sync_with_stdio(0);
  init();
  preprocess();
  solve();
  return 0;
}
