#include <bits/stdc++.h>
using namespace std;
int arr[1000000], is[4001][4001], memo[4001][4001];
map<int, set<int> > mp;
int recursive(int s1, int s2) {
  if (is[s1][s2]) return memo[s1][s2];
  is[s1][s2] = true;
  if (mp[arr[s1]].upper_bound(s2) != mp[arr[s1]].end()) {
    int yo = *mp[arr[s1]].upper_bound(s2);
    return memo[s1][s2] = recursive(s2, yo) + 1;
  } else
    return memo[s1][s2] = 0;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    mp[arr[i]].insert(i);
  }
  int maxi = 1;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      maxi = max(maxi, recursive(i, j) + 2);
    }
  }
  cout << maxi;
  return 0;
}

### Prompt

Your challenge is to write 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;
int arr[1000000], is[4001][4001], memo[4001][4001];
map<int, set<int> > mp;
int recursive(int s1, int s2) {
  if (is[s1][s2]) return memo[s1][s2];
  is[s1][s2] = true;
  if (mp[arr[s1]].upper_bound(s2) != mp[arr[s1]].end()) {
    int yo = *mp[arr[s1]].upper_bound(s2);
    return memo[s1][s2] = recursive(s2, yo) + 1;
  } else
    return memo[s1][s2] = 0;
}
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    mp[arr[i]].insert(i);
  }
  int maxi = 1;
  for (int i = 0; i < n; i++) {
    for (int j = i + 1; j < n; j++) {
      maxi = max(maxi, recursive(i, j) + 2);
    }
  }
  cout << maxi;
  return 0;
}

```