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;
const int maxn = 1000 + 85 - 69;
int a[maxn], cnt[maxn], n;
vector<int> all;
int f(int x, int y) {
  int z = x + y, ret;
  int idz = lower_bound(all.begin(), all.end(), z) - all.begin();
  if (idz != ((int)(all).size()) and all[idz] == z and cnt[idz])
    cnt[idz]--, ret = 1 + f(y, z), cnt[idz]++;
  else
    ret = 2;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    all.push_back(a[i]);
  }
  sort(all.begin(), all.end());
  all.resize(unique(all.begin(), all.end()) - all.begin());
  for (int i = 0; i < n; i++)
    cnt[lower_bound(all.begin(), all.end(), a[i]) - all.begin()]++;
  int ans = 2;
  for (int i = 0; i < ((int)(all).size()); i++) {
    cnt[i]--;
    for (int j = 0; j < ((int)(all).size()); j++)
      if (cnt[j]) cnt[j]--, ((ans) = max((ans), (f(all[i], all[j])))), cnt[j]++;
    cnt[i]++;
  }
  cout << ans << endl;
  return 0;
}
