#include <bits/stdc++.h>
using namespace std;
const int MAX = 4010;
int arr[MAX];
int temp[MAX];
int memo[MAX][MAX];
vector<int> idx[MAX];
map<int, int> mp;
int n;
int solve(int i, int prev) {
  int &ret = memo[i][prev];
  if (~ret) return ret;
  ret = -1e9;
  int next = arr[prev];
  auto it = upper_bound(idx[next].begin(), idx[next].end(), i);
  if (it != idx[next].end()) {
    return ret = max(ret, solve(*it, i) + 1);
  } else
    return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &arr[i]), temp[i] = arr[i];
  if (n <= 2) {
    printf("%d\n", n);
    return 0;
  }
  sort(temp, temp + n);
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    if (mp.count(temp[i]) == 0) mp[temp[i]] = ++cnt;
  }
  for (int i = 0; i < n; ++i) arr[i] = mp[arr[i]];
  for (int i = 0; i < n; ++i) idx[arr[i]].push_back(i);
  memset(memo, -1, sizeof(memo));
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < i; ++j) {
      ans = max(ans, solve(i, j));
    }
  }
  printf("%d\n", ans + 2);
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 MAX = 4010;
int arr[MAX];
int temp[MAX];
int memo[MAX][MAX];
vector<int> idx[MAX];
map<int, int> mp;
int n;
int solve(int i, int prev) {
  int &ret = memo[i][prev];
  if (~ret) return ret;
  ret = -1e9;
  int next = arr[prev];
  auto it = upper_bound(idx[next].begin(), idx[next].end(), i);
  if (it != idx[next].end()) {
    return ret = max(ret, solve(*it, i) + 1);
  } else
    return 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", &arr[i]), temp[i] = arr[i];
  if (n <= 2) {
    printf("%d\n", n);
    return 0;
  }
  sort(temp, temp + n);
  int cnt = 0;
  for (int i = 0; i < n; ++i) {
    if (mp.count(temp[i]) == 0) mp[temp[i]] = ++cnt;
  }
  for (int i = 0; i < n; ++i) arr[i] = mp[arr[i]];
  for (int i = 0; i < n; ++i) idx[arr[i]].push_back(i);
  memset(memo, -1, sizeof(memo));
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < i; ++j) {
      ans = max(ans, solve(i, j));
    }
  }
  printf("%d\n", ans + 2);
}

```