Yash has recently learnt about the Fibonacci sequence and is very excited about it. He calls a sequence Fibonacci-ish if 

  1. the sequence consists of at least two elements 
  2. f0 and f1 are arbitrary 
  3. fn + 2 = fn + 1 + fn for all n ≥ 0. 



You are given some sequence of integers a1, a2, ..., an. Your task is rearrange elements of this sequence in such a way that its longest possible prefix is Fibonacci-ish sequence.

Input

The first line of the input contains a single integer n (2 ≤ n ≤ 1000) — the length of the sequence ai.

The second line contains n integers a1, a2, ..., an (|ai| ≤ 109).

Output

Print the length of the longest possible Fibonacci-ish prefix of the given sequence after rearrangement.

Examples

Input

3
1 2 -1


Output

3


Input

5
28 35 7 14 21


Output

4

Note

In the first sample, if we rearrange elements of the sequence as  - 1, 2, 1, the whole sequence ai would be Fibonacci-ish.

In the second sample, the optimal way to rearrange elements is <image>, <image>, <image>, <image>, 28.
// practice with rainboy
import java.io.*;
import java.util.*;

public class CF633D extends PrintWriter {
	CF633D() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF633D o = new CF633D(); o.main(); o.flush();
	}

	Map<Integer, Integer> mp;
	int solve(int a_, int b_) {
		mp.put(a_, mp.get(a_) - 1);
		mp.put(b_, mp.get(b_) - 1);
		int m = 2;
		for (int a = a_, b = b_, c; ; a = b, b = c) {
			c = a + b;
			int k = mp.getOrDefault(c, 0);
			if (k == 0) {
				mp.put(a_, mp.get(a_) + 1);
				mp.put(b_, mp.get(b_) + 1);
				a = a_; b = b_;
				for (int i = 2; i < m; i++) {
					c = a + b;
					mp.put(c, mp.get(c) + 1);
					a = b; b = c;
				}
				return m;
			}
			mp.put(c, k - 1);
			m++;
		}
	}
	void main() {
		int n = sc.nextInt();
		int[] aa = new int[n];
		mp = new HashMap<>();
		for (int i = 0; i < n; i++) {
			aa[i] = sc.nextInt();
			mp.put(aa[i], mp.getOrDefault(aa[i], 0) + 1);
		}
		int ans = mp.getOrDefault(0, 0);
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++)
				if (i != j && !(aa[i] == 0 && aa[j] == 0))
					ans = Math.max(ans, solve(aa[i], aa[j]));
		println(ans);
	}
}
