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_base::sync_with_stdio(false);
  cout << fixed << setprecision(7);
  int n;
  std::cin >> n;
  std::vector<long long> a(n);
  int nuls = 0;
  for (size_t i = 0; i < n; i++) {
    std::cin >> a[i];
    if (a[i] == 0) {
      ++nuls;
    }
  }
  int bestans = max(2, nuls);
  if (nuls > 10) {
    std::vector<long long> b;
    for (size_t i = 0; i < n; i++) {
      if (a[i] != 0) {
        b.push_back(a[i]);
      }
    }
    b.insert(b.end(), 10, 0);
    a = b;
  }
  n = a.size();
  unordered_multiset<long long> base(a.begin(), a.end());
  set<pair<long long, long long>> was;
  for (size_t i = 0; i < n; i++) {
    for (size_t j = 0; j < n; j++) {
      if (i == j || was.find(make_pair(a[i], a[j])) != was.end()) {
        continue;
      }
      was.insert(make_pair(a[i], a[j]));
      std::vector<long long> del{a[i], a[j]};
      base.erase(base.find(a[i]));
      base.erase(base.find(a[j]));
      auto f = a[i];
      auto s = a[j];
      int ans = 2;
      for (size_t k = 0; k < n; k++) {
        auto t = f + s;
        auto it = base.find(t);
        if (it == base.end()) {
          break;
        } else {
          ++ans;
          del.push_back(*it);
          base.erase(it);
        }
        f = s;
        s = t;
      }
      for (auto&& element : del) {
        base.insert(element);
      }
      bestans = max(ans, bestans);
    }
  }
  std::cout << bestans << std::endl;
  return 0;
}
