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;
const long long int mod = 1000000007;
map<long long int, long long int> counts;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int n;
  cin >> n;
  long long int arr[n];
  for (long long int i = 0; i < n; i++) {
    cin >> arr[i];
    counts[arr[i]] = 0;
  }
  for (long long int i = 0; i < n; i++) counts[arr[i]]++;
  long long int ans = 0, a, b, c;
  for (long long int i = 0; i < n; i++) {
    for (long long int j = 0; j < n; j++) {
      if (i == j) continue;
      a = arr[i], b = arr[j];
      if (!a and !b) {
        ans = max(ans, counts[a]);
        continue;
      }
      vector<long long int> repair;
      repair.push_back(a);
      repair.push_back(b);
      counts[a]--;
      counts[b]--;
      while (1) {
        c = a + b;
        if (counts.find(c) != counts.end() and counts[c] > 0) {
          counts[c]--;
          repair.push_back(c);
          a = b;
          b = c;
        } else
          break;
      }
      ans = max(ans, (long long int)repair.size());
      for (auto it : repair) counts[it]++;
    }
  }
  cout << ans << endl;
  return 0;
}
