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;
long long arr[10009 + 5];
map<long long, int> m1;
long long n;
long long calc(long long a, long long b) {
  long long ct = 0;
  ct += 2;
  long long tmp;
  vector<long long> v;
  m1[a]--;
  m1[b]--;
  v.push_back(a);
  v.push_back(b);
  while (1) {
    tmp = a + b;
    if (m1[tmp] == 0) break;
    ct++;
    v.push_back(tmp);
    m1[tmp]--;
    if (ct == n) break;
    a = b;
    b = tmp;
  }
  for (int i = 0; i < v.size(); i++) {
    m1[v[i]]++;
  }
  return ct;
}
int main() {
  cin >> n;
  map<pair<long long, long long>, int> mpair;
  m1.clear();
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    m1[arr[i]]++;
  }
  long long mx = -1, flg = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      if (mpair[make_pair(arr[i], arr[j])]) continue;
      mpair[make_pair(arr[i], arr[j])] = 1;
      mx = max(mx, calc(arr[i], arr[j]));
      if (mx == n) {
        flg = 1;
        break;
      }
    }
    if (flg) break;
  }
  cout << mx << endl;
}
