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> mp;
int a[1005];
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> a[i], mp[a[i]]++;
  sort(a, a + n);
  int ans = 0;
  vector<int> v;
  if (mp[0] >= 2) ans = max(ans, mp[0]);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i != j) {
        v.clear();
        if (a[i] == 0 && a[j] == 0) continue;
        int x = a[i], y = a[j];
        int l = 0;
        while (true) {
          if (mp[x] == 0) break;
          l++;
          mp[x]--;
          v.push_back(x);
          int z = x + y;
          x = y, y = z;
        }
        for (int z = 0; z < v.size(); z++) mp[v[z]]++;
        ans = max(ans, l);
      }
    }
  }
  cout << ans << endl;
}
