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>
#pragma GCC optimize("Ofast")
using namespace std;
using ll = int;
int main() {
  ios_base::sync_with_stdio(0);
  cout.tie(0);
  cin.tie(0);
  ll n;
  cin >> n;
  vector<ll> a(n);
  for (ll i = 0; i < n; i++) {
    cin >> a[i];
  }
  map<ll, ll> c;
  for (ll i = 0; i < n; i++) {
    c[a[i]]++;
  }
  ll len = 0;
  for (ll i = 0; i < n; i++) {
    for (ll j = 0; j < n; j++) {
      if (i == j) {
        continue;
      }
      if (a[i] == 0 && a[j] == 0) {
        continue;
      }
      vector<ll> fib;
      fib.push_back(a[i]);
      fib.push_back(a[j]);
      c[a[i]]--;
      c[a[j]]--;
      while (c[fib[ll(fib.size()) - 1] + fib[ll(fib.size()) - 2]] > 0) {
        fib.push_back(fib[ll(fib.size()) - 1] + fib[ll(fib.size()) - 2]);
        c[fib[ll(fib.size()) - 1]]--;
      }
      len = max(len, ll(fib.size()));
      for (auto k : fib) {
        c[k]++;
      }
    }
  }
  len = max(len, c[0]);
  cout << len;
  return 0;
}
