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.
n = int(input())
a = list(map(int,input().split()))
s = {}			#dic
d = {}			#dic
p = set()		#set
mx = - 10**15
mi = -mx
ans = 2
for i in range(n) :
	if a[i] in s :
		s[a[i]] += 1
	else :
		s[a[i]] = 1
	mx = max(mx,a[i])
	mi = min(mi,a[i])
if 0 in s : ans = max(ans,s[0])
for i in range(n) :
	for j in range(n) :
		if (j==i) or (a[i]==0 and a[j]==0) or ((a[i], a[j]) in p) : continue
		f0 = a[i]; f1 = a[j]; ansi = 2; f2 = 0;
		d = {}
		p.add((a[i], a[j]))
		while True :
			f2 = f1 + f0
			f0 = f1
			f1 = f2
			if (f2 > mx) or (f2 < mi) : break
			if f2 in s :
				d[f2] = s[f2]
			else : break
		f0 = a[i]; f1 = a[j];
		if f1==f0 :
			d[f0] = s[f0] - 2
		else :
			d[f0] = s[f0] - 1
			d[f1] = s[f1] - 1
		while True :
			f2 = f1 + f0
			f0 = f1
			f1 = f2
			if (f2 > mx) or (f2 < mi) : break
			if (f2 in d) and (d[f2] > 0) :
				ansi += 1
				d[f2] -= 1
			else : break
		ans = max(ans, ansi)
print(ans)

