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.
import java.util.*;
public class Main {
	public static void main(String[] args) {
		Scanner stdin = new Scanner(System.in);
		long CC = (long)1e10 + 7;
		int n = stdin.nextInt();
		long[] a = new long[n];
		HashSet<Long> set = new HashSet<>();
		HashMap<Long, Integer> map = new HashMap<>();
		
		for (int i = 0; i < n; i++) {
			a[i] = stdin.nextLong();
			map.put(a[i], map.containsKey(a[i]) ? map.get(a[i]) + 1 : 1);
		}
		
		int ans = 2;
		for (int i = 0; i < n; i++)
			for (int j = 0; j < n; j++) {
				if (i == j) continue;
				long KEY = CC * a[i] + a[j];
				
				if (set.contains(KEY)) continue;
				set.add(KEY);
				
				long a1 = a[i];
				long a2 = a[j];
				map.put(a[i], map.get(a[i]) - 1);
				map.put(a[j], map.get(a[j]) - 1);
				
				int tmp = 2;
				
				for(;;) {
					long a3 = a1 + a2;
					if (map.containsKey(a3) && map.get(a3) > 0) {
						tmp++;
						map.put(a3, map.get(a3) - 1);
						a1 = a2;
						a2 = a3;
					} else
						break;
				}
				
				ans = Math.max(ans, tmp);
				a1 = a[i];
				a2 = a[j];
				map.put(a[i], map.get(a[i]) + 1);
				map.put(a[j], map.get(a[j]) + 1);
				for (int k = 2; k < tmp; k++) {
					long a3 = a1 + a2;
					map.put(a3, map.get(a3) + 1);
					a1 = a2;
					a2 = a3;
				}
			}
		System.out.println(ans);
	}
}
