Arithmetic Progressions

An arithmetic progression is a sequence of numbers $a_1, a_2, ..., a_k$ where the difference of consecutive members $a_{i+1} - a_i$ is a constant ($1 \leq i \leq k-1$). For example, the sequence 5, 8, 11, 14, 17 is an arithmetic progression of length 5 with the common difference 3.

In this problem, you are requested to find the longest arithmetic progression which can be formed selecting some numbers from a given set of numbers. For example, if the given set of numbers is {0, 1, 3, 5, 6, 9}, you can form arithmetic progressions such as 0, 3, 6, 9 with the common difference 3, or 9, 5, 1 with the common difference -4. In this case, the progressions 0, 3, 6, 9 and 9, 6, 3, 0 are the longest.

Input

The input consists of a single test case of the following format.


$n$
$v_1$ $v_2$ ... $v_n$


$n$ is the number of elements of the set, which is an integer satisfying $2 \leq n \leq 5000$. Each $v_i$ ($1 \leq i \leq n$) is an element of the set, which is an integer satisfying $0 \leq v_i \leq 10^9$. $v_i$'s are all different, i.e., $v_i \ne v_j$ if $i \ne j$.

Output

Output the length of the longest arithmetic progressions which can be formed selecting some numbers from the given set of numbers.

Sample Input 1


6
0 1 3 5 6 9


Sample Output 1


4


Sample Input 2


7
1 4 7 3 2 6 5


Sample Output 2


7


Sample Input 3


5
1 2 4 8 16


Sample Output 3


2






Example

Input

6
0 1 3 5 6 9


Output

4
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;

int n[5010];
int dp[5010][5010];

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
#ifdef LOCAL
	std::ifstream in("in");
	std::cin.rdbuf(in.rdbuf());
#endif

	int N;
	cin >> N;
	for(int i = 0; i < N; i++){
		cin >> n[i];
	};
	sort(n, n + N);

	memset(dp, -1, sizeof dp);
	for(int i = 0; i < N; i++){
		for(int j = 0; j < i; j++){
			dp[i][j] = 2;
		}
	}
	for(int i = 0; i < N; i++){
		for(int j = 0; j < i; j++){
			if(dp[i][j] == -1) continue;
			int pre = n[j];
			int d = n[i] - pre;
			int idx = lower_bound(n, n + N, n[i] + d) - n;
			if(idx < N && n[i] + d == n[idx]){
				dp[idx][i] = max(dp[idx][i], dp[i][j] + 1);
			}
		}
	}

	int ans = 0;
	for(int i = 0; i < N; i++){
		for(int j = 0; j < N; j++){
			ans = max(ans, dp[i][j]);
		}
	}
	cout << ans << endl;
}

