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 N = 1e3 + 5;
int a[N];
map<int, vector<int> > indx;
map<pair<int, int>, bool> vis;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    indx[a[i]].push_back(i);
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      vector<int> vec;
      int f = a[i], s = a[j];
      if (vis[make_pair(f, s)]) continue;
      vis[make_pair(f, s)] = 1;
      vec.push_back(i), vec.push_back(j);
      indx[f].pop_back(), indx[s].pop_back();
      for (int k = 0; k < n; k++) {
        int temp = s + f;
        if (indx[temp].size()) {
          vec.push_back(indx[temp].back()), indx[temp].pop_back();
          f = s, s = temp;
        } else
          break;
      }
      ans = max(ans, (int)vec.size());
      for (int k = 0; k < vec.size(); k++) indx[a[vec[k]]].push_back(vec[k]);
    }
  }
  cout << ans << endl;
  return 0;
}
