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;
using ll = long long;
using ld = long double;
using ii = pair<ll, ll>;
using vi = vector<ll>;
using vb = vector<bool>;
using vvi = vector<vi>;
using vii = vector<ii>;
using vvii = vector<vii>;
const int INF = 2000000000;
const ll LLINF = 9000000000000000000;
void add(map<ll, int> &m, ll v) {
  auto it = m.find(v);
  if (it == m.end())
    m.insert({v, 1});
  else
    it->second++;
}
void remove(map<ll, int> &m, ll v) {
  auto it = m.find(v);
  if (it->second > 1)
    it->second--;
  else
    m.erase(it);
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int N;
  cin >> N;
  vi A(N, 0LL);
  for (int i = 0; i < N; ++i) cin >> A[i];
  sort(A.begin(), A.end());
  set<ii> had;
  map<ll, int> rem;
  for (ll v : A) add(rem, v);
  vi q;
  ll L = 2LL;
  for (int i = 0; i < N; ++i)
    for (int j = 0; j < N; ++j) {
      if (i == j) continue;
      if (had.find({A[i], A[j]}) != had.end()) continue;
      q.push_back(A[i]);
      q.push_back(A[j]);
      remove(rem, A[i]);
      remove(rem, A[j]);
      had.insert({A[i], A[j]});
      while (rem.size() > 0) {
        ll nxt = q.back() + q[q.size() - 2];
        if (rem.find(nxt) == rem.end()) break;
        remove(rem, nxt);
        q.push_back(nxt);
      }
      if (q.size() > L) {
        L = q.size();
      }
      while (q.size() > 0) {
        add(rem, q.back());
        q.pop_back();
      }
    }
  cout << L << endl;
  return 0;
}
