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[1009], ans, ma, mi = -2e9;
map<int, int> m;
int calc(int i, int j) {
  int t = 2;
  if (1LL * i + j > ma || 1LL * i + j < mi) return t;
  int p = j, q = i + j;
  vector<int> v;
  m[i]--;
  m[j]--;
  v.push_back(i);
  v.push_back(j);
  while (1) {
    if (m.find(q) == m.end() || m[q] == 0) break;
    t++;
    m[q]--;
    v.push_back(q);
    q += p;
    p = q - p;
  }
  for (int i = 0; i < v.size(); i++) m[v[i]]++;
  return t;
}
int main() {
  cin >> n;
  int z = 0;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    m[a[i]]++;
    if (!a[i]) z++;
    ma = max(ma, a[i]);
    mi = min(a[i], mi);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      if (!a[i] && !a[j]) {
        continue;
      }
      int t = calc(a[i], a[j]);
      ans = max(ans, calc(a[i], a[j]));
    }
  }
  cout << max(z, ans);
  return 0;
}
