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 long long maxn = 1000 + 10;
long long a[maxn], n;
set<pair<long long, long long>> s;
unordered_map<long long, long long> m;
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (long long i = 0; i < n; ++i) cin >> a[i];
  long long ans = 0;
  for (long long i = 0; i < n; ++i) ++m[a[i]];
  for (long long i = 0; i < n; ++i) {
    for (long long j = 0; j < n; ++j)
      if (i != j) {
        if (s.count(make_pair(a[i], a[j]))) continue;
        vector<long long> f = {a[i], a[j]};
        --m[a[i]];
        --m[a[j]];
        while (m[f[f.size() - 1] + f[f.size() - 2]]) {
          --m[f[f.size() - 1] + f[f.size() - 2]];
          f.push_back(f[f.size() - 1] + f[f.size() - 2]);
        }
        s.insert(make_pair(a[i], a[j]));
        ans = max(ans, (long long)f.size());
        for (long long k : f) ++m[k];
      }
  }
  cout << ans << endl;
  return 0;
}
