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.
#include <bits/stdc++.h>
using namespace std;
int a[1003], ans[1003];
queue<int> q;
map<int, int> m;
int main() {
  int n, ans = 0, x, y;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]), ++m[a[i]];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      if (a[i] == a[j] && a[i] == 0) {
        ans = max(ans, m[0]);
        continue;
      }
      int k = 0;
      x = a[i], y = a[j], --m[a[i]], --m[a[j]], q.push(a[i]), q.push(a[j]);
      while (m[x + y]) --m[x + y], q.push(x + y), y = x + y, x = y - x, ++k;
      ans = max(ans, k + 2);
      while (!q.empty()) ++m[q.front()], q.pop();
    }
  printf("%d", ans);
  return 0;
}
