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 n, a[1005], ans, x, y, z, ln;
unordered_map<int, int> mp;
int elements[1005];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    ++mp[a[i]];
  }
  if (mp.count(0))
    ans = max(2, mp[0]);
  else
    ans = 2;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (i == j) continue;
      if (a[i] == 0 && a[j] == 0) continue;
      ln = 2;
      x = a[i], y = a[j];
      elements[1] = x, elements[2] = y;
      --mp[x], --mp[y];
      while (1) {
        z = x + y;
        if (!mp[z]) break;
        elements[++ln] = z;
        --mp[z];
        x = y, y = z;
      }
      for (int k = 1; k <= ln; ++k) ++mp[elements[k]];
      ans = max(ln, ans);
    }
  }
  cout << ans << '\n';
  return 0;
}
