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>
const int MAX = 1000000000;
int64_t a[1000];
std::unordered_map<int, int> cnt;
int64_t F[400];
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  std::cin >> n;
  cnt.reserve(n);
  cnt.max_load_factor(0.25);
  for (int i = 0; i < n; ++i) {
    std::cin >> a[i];
    cnt[a[i]]++;
  }
  int ans = std::max(2, cnt[0]);
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      if (i == j) continue;
      if (a[i] == 0 && a[j] == 0) continue;
      F[0] = a[i];
      cnt[a[i]]--;
      F[1] = a[j];
      cnt[a[j]]--;
      int len = 2;
      while (std::abs(F[len - 1] + F[len - 2]) <= MAX) {
        F[len] = F[len - 1] + F[len - 2];
        if (cnt[F[len]] > 0) {
          cnt[F[len]]--;
          len++;
        } else {
          break;
        }
      }
      for (int k = 0; k < len; ++k) {
        cnt[F[k]]++;
      }
      ans = std::max(ans, len);
    }
  }
  std::cout << ans << '\n';
  return 0;
}
