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;
map<int, int> Map;
int A[1111];
int n;
int main() {
  cin >> n;
  for (int i = 1; i <= n; ++i) scanf("%d", &A[i]), ++Map[A[i]];
  sort(A + 1, A + n + 1);
  if (A[0] == A[1]) --A[0];
  int ans = 1;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      if ((i != j && A[i] != A[i - 1] && A[j] != A[j - 1]) ||
          (A[i] == A[j] && A[i] != A[i - 1] && j == i + 1)) {
        --Map[A[i]];
        --Map[A[j]];
        int x = A[i], y = A[j], len = 2;
        vector<int> record;
        while (1) {
          if ((long long)x + y == x + y && Map.count(x + y) && Map[x + y] > 0) {
            ++len;
          } else
            break;
          int t = x + y;
          x = y;
          y = t;
          --Map[y];
          record.push_back(t);
        }
        ans = max(ans, len);
        for (auto e : record) ++Map[e];
        ++Map[A[i]];
        ++Map[A[j]];
      }
  cout << ans << endl;
  return 0;
}
