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>
#pragma GCC optimize("-Ofast", "-funroll-all-loops")
using namespace std;
const int N = 1e3 + 10;
int n, a[N], res;
map<int, int> mp;
signed main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i], mp[a[i]]++;
  res = max(res, mp[0]);
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (i != j) {
        if (!a[i] && !a[j]) continue;
        int mx = 2, now = a[i] + a[j], pre = a[j];
        map<int, int> tmp;
        tmp[a[i]] = 1, tmp[a[j]] = 1;
        while (tmp[now] < mp[now]) {
          tmp[now]++;
          int f = now;
          now = now + pre;
          pre = f;
          mx++;
        }
        res = max(res, mx);
      }
  cout << res;
  return 0;
}
