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;
vector<int> v;
int n;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  v.resize(n);
  map<int, int> mp;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    mp[v[i]]++;
  }
  sort(v.begin(), v.end());
  auto it = unique(v.begin(), v.end());
  v.resize(distance(v.begin(), it));
  int sz = (int)v.size();
  int ans = 2ll;
  ans = max(ans, mp[0]);
  for (int i = 0; i < sz; i++) {
    for (int j = 0; j < sz; j++) {
      if ((v[i] == 0 and v[j] == 0) or (i == j and mp[v[i]] == 1)) continue;
      int a = v[i], b = v[j];
      int cur = 2;
      map<int, int> ump;
      ump[a]++;
      ump[b]++;
      while (true) {
        if (mp.count(a + b) and ump[a + b] < mp[a + b]) {
          ump[a + b]++;
          a = a + b;
          swap(a, b);
          cur++;
        } else
          break;
      }
      ans = max(ans, cur);
    }
  }
  cout << ans << endl;
}
