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>
const int dx[] = {1, 0, -1, 0};
const int dy[] = {0, 1, 0, -1};
using namespace std;
int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  vector<int> a(n);
  unordered_map<long long, int> mp;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mp[a[i]]++;
  }
  int ans = mp[0];
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i != j) {
        if (a[i] == 0 && a[j] == 0) continue;
        vector<long long> V;
        V.push_back(a[i]);
        V.push_back(a[j]);
        mp[a[i]]--;
        mp[a[j]]--;
        while (1) {
          int len = V.size();
          long long v = V[len - 1] + V[len - 2];
          if (mp.find(v) == mp.end() || mp[v] == 0) {
            ans = max(ans, len);
            break;
          }
          mp[v]--;
          V.push_back(v);
        }
        for (long long el : V) mp[el]++;
      }
  cout << ans << endl;
  return 0;
}
