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 main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int n;
  cin >> n;
  vector<int> a(n);
  map<int, int> m;
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    ++m[a[i]];
  }
  int ans = 2;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      if (i != j) {
        if (!a[i] && !a[j]) {
          ans = max(ans, m[0]);
          continue;
        }
        vector<int> f(2);
        f[0] = a[i], f[1] = a[j];
        --m[f[0]], --m[f[1]];
        while (auto &cnt = m[f[f.size() - 1] + f[f.size() - 2]]) {
          f.push_back(f[f.size() - 1] + f[f.size() - 2]);
          --cnt;
        }
        ans = max(ans, (int)f.size());
        for (int i : f) {
          ++m[i];
        }
      }
    }
  }
  cout << ans << endl;
  return 0;
}
