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[100000], m[100000];
map<int, int> c;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, ans = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    c[a[i]]++;
    if (a[i] == 0) ans++;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i != j && (a[i] || a[j])) {
        m[0] = a[i];
        c[a[i]]--;
        m[1] = a[j];
        c[a[j]]--;
        int t = 2;
        while (1) {
          m[t] = m[t - 1] + m[t - 2];
          if (c[m[t]]) {
            c[m[t]]--;
            t++;
          } else
            break;
        }
        ans = max(ans, t);
        for (--t; t >= 0; t--) {
          c[m[t]]++;
        }
      }
    }
  }
  cout << ans;
  return 0;
}
