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() {
  int n;
  cin >> n;
  vector<int> nums(n);
  for (int i = 0; i < n; i++) cin >> nums[i];
  map<int, int> mp;
  for (int i : nums) mp[i]++;
  int ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      int f0 = nums[i], f1 = nums[j], c = 0;
      if (f0 == 0 && f1 == 0) {
        c = mp[0];
        ans = max(ans, c);
        break;
      }
      mp[f0]--;
      mp[f1]--;
      while (mp.count(f0 + f1) && mp[f0 + f1] > 0) {
        mp[f0 + f1]--;
        int k = f0;
        f0 = f1;
        f1 = k + f1;
        c++;
      }
      ans = max(ans, c + 2);
      mp[f0]++;
      mp[f1]++;
      while (c-- > 0) {
        int f = f1 - f0;
        mp[f]++;
        f1 = f0;
        f0 = f;
      }
    }
  }
  cout << ans;
}
