#include <bits/stdc++.h>
using namespace std;
const int MAXN = 4000 + 10;
int dp[MAXN][MAXN], a[MAXN];
vector<int> last[1000 * 1000 + 10];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) dp[i][i] = 1;
  int ans = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < i; j++) {
      int k = upper_bound(last[a[i]].begin(), last[a[i]].end(), j) -
              last[a[i]].begin();
      k--;
      if (k >= 0 and last[a[i]][k] == j) k--;
      if (k < 0 or last[a[i]][k] > j)
        dp[i][j] = 2;
      else {
        k = last[a[i]][k];
        dp[i][j] = 1 + dp[j][k];
      }
      ans = max(ans, dp[i][j]);
    }
    last[a[i]].push_back(i);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 4000 + 10;
int dp[MAXN][MAXN], a[MAXN];
vector<int> last[1000 * 1000 + 10];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = 0; i < n; i++) dp[i][i] = 1;
  int ans = 1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < i; j++) {
      int k = upper_bound(last[a[i]].begin(), last[a[i]].end(), j) -
              last[a[i]].begin();
      k--;
      if (k >= 0 and last[a[i]][k] == j) k--;
      if (k < 0 or last[a[i]][k] > j)
        dp[i][j] = 2;
      else {
        k = last[a[i]][k];
        dp[i][j] = 1 + dp[j][k];
      }
      ans = max(ans, dp[i][j]);
    }
    last[a[i]].push_back(i);
  }
  cout << ans << endl;
  return 0;
}

```