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<long long, long long> m;
long long a[10004];
long long f(long long a, long long b) {
  m[a]--;
  m[b]--;
  vector<long long> v;
  v.push_back(a);
  v.push_back(b);
  long long x = b;
  long long y = a + b;
  while (m[y] != 0) {
    m[y]--;
    v.push_back(y);
    y = x + y;
    x = y - x;
  }
  for (long long i = 0; i < v.size(); i++) m[v[i]]++;
  return v.size();
}
signed main() {
  long long n, ans;
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    m[a[i]]++;
  }
  ans = -1;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      if (i != j) {
        if (a[i] == a[j] && !a[i])
          ans = max(ans, m[0]);
        else {
          ans = max(ans, f(a[i], a[j]));
        }
      }
    }
  }
  cout << ans << endl;
}
