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.io.*;
import java.util.*;

public class D {

	BufferedReader br;
	PrintWriter out;
	StringTokenizer st;
	boolean eof;

	void solve() throws IOException {
		int n = nextInt();
		int[] a = new int[n];
		for (int i = 0; i < n; i++) {
			a[i] = nextInt();
		}

		Arrays.sort(a);
		int sz = 1;
		int[] cnt = new int[n];
		cnt[0] = 1;
		for (int i = 1; i < n; i++) {
			if (a[i] != a[sz - 1]) {
				a[sz++] = a[i];
			}
			cnt[sz - 1]++;
		}

		a = Arrays.copyOf(a, sz);
		cnt = Arrays.copyOf(cnt, sz);

		int ans = 2;

		int[] last = new int[sz];
		int[] tmpCnt = new int[sz];
		int time = 0;
		for (int i = 0; i < sz; i++)
			for (int j = 0; j < sz; j++) {
				if (i == j && cnt[i] < 2) {
					continue;
				}
				time++;
				
				tmpCnt[i] = cnt[i];
				tmpCnt[j] = cnt[j];
				last[i] = time;
				last[j] = time;
				tmpCnt[i]--;
				tmpCnt[j]--;

				int x = a[i];
				int y = a[j];

				int len = 2;

				while (true) {
					int newY = Arrays.binarySearch(a, x + y);
					if (newY < 0) {
						break;
					}
					if (last[newY] != time) {
						last[newY] = time;
						tmpCnt[newY] = cnt[newY];
					}
					if (tmpCnt[newY] == 0) {
						break;
					}
					tmpCnt[newY]--;
					len++;
					newY = x + y;
					x = y;
					y = newY;
				}

//				System.err.println(i + " " + j + " " + len);

				ans = Math.max(ans, len);
			}

		out.println(ans);

	}

	D() throws IOException {
		br = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		solve();
		out.close();
	}

	public static void main(String[] args) throws IOException {
		new D();
	}

	String nextToken() {
		while (st == null || !st.hasMoreTokens()) {
			try {
				st = new StringTokenizer(br.readLine());
			} catch (Exception e) {
				eof = true;
				return null;
			}
		}
		return st.nextToken();
	}

	String nextString() {
		try {
			return br.readLine();
		} catch (IOException e) {
			eof = true;
			return null;
		}
	}

	int nextInt() throws IOException {
		return Integer.parseInt(nextToken());
	}

	long nextLong() throws IOException {
		return Long.parseLong(nextToken());
	}

	double nextDouble() throws IOException {
		return Double.parseDouble(nextToken());
	}
}