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 a[1234], n;
multiset<int> mlt;
int guess(int x, int y) {
  vector<int> used;
  used.push_back(x);
  pair<multiset<int>::iterator, multiset<int>::iterator> px =
      mlt.equal_range(x);
  mlt.erase(px.first);
  used.push_back(y);
  pair<multiset<int>::iterator, multiset<int>::iterator> py =
      mlt.equal_range(y);
  mlt.erase(py.first);
  int tmp;
  while (1) {
    tmp = x + y;
    pair<multiset<int>::iterator, multiset<int>::iterator> pit =
        mlt.equal_range(tmp);
    if (pit.first != mlt.end()) {
      if (*pit.first != tmp) break;
      x = y;
      y = tmp;
      mlt.erase(pit.first);
      used.push_back(y);
    } else {
      break;
    }
  }
  for (auto i : used) {
    mlt.insert(i);
  }
  return used.size();
}
int main() {
  cin >> n;
  map<int, int> used;
  for (int i(0); i < n; i++) {
    cin >> a[i];
    if (!used.count(a[i]))
      used[a[i]] = 1;
    else
      used[a[i]] += 1;
  }
  int mx = 0;
  for (int i(0); i < n; i++) mlt.insert(a[i]);
  for (auto i : used) {
    for (auto j : used) {
      if (i.first == j.first)
        if (i.second < 2) continue;
      if (i.first == 0 && j.first == 0)
        mx = max(mx, used[0]);
      else
        mx = max(mx, guess(i.first, j.first));
    }
  }
  cout << mx << endl;
  return 0;
}
