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>
int dfs(int a, int b, std::map<int, int>& cnt) {
  if (cnt[a + b] != 0) {
    cnt[a + b]--;
    int ret = 1 + dfs(b, a + b, cnt);
    cnt[a + b]++;
    return ret;
  }
  return 0;
}
int main() {
  int numbers;
  std::scanf("%d", &numbers);
  std::vector<int> values;
  std::map<int, int> cnt;
  for (int i = 0; i < numbers; i++) {
    int next;
    std::scanf("%d", &next);
    values.push_back(next);
    cnt[next]++;
  }
  constexpr int MAXN = 1000;
  int answer = 0;
  std::set<std::pair<int, int>> done;
  for (int i = 0; i < numbers; i++) {
    for (int j = 0; j < numbers; j++) {
      if (i == j) continue;
      if (done.find({values[i], values[j]}) != done.end()) continue;
      cnt[values[i]]--;
      cnt[values[j]]--;
      answer = std::max(answer, 2 + dfs(values[i], values[j], cnt));
      done.insert({values[i], values[j]});
      cnt[values[i]]++;
      cnt[values[j]]++;
    }
  }
  std::cout << answer << std::endl;
  return EXIT_SUCCESS;
}
