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 n, a[1111];
map<int, int> m;
map<int, int>::iterator it;
vector<int> v;
int main() {
  while (~scanf("%d", &n)) {
    m.clear();
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      m[a[i]]++;
    }
    sort(a + 1, a + n + 1);
    int ans = m[0];
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++)
        if (i != j && (a[i] != 0 || a[j] != 0)) {
          v.clear();
          v.push_back(a[i]), v.push_back(a[j]);
          m[a[i]]--, m[a[j]]--;
          int f0 = a[i], f1 = a[j], num = 2;
          while (1) {
            if (m[f0 + f1]) {
              m[f0 + f1]--, v.push_back(f0 + f1);
              num++;
              swap(f0, f1), f1 = f0 + f1;
            } else
              break;
          }
          ans = max(ans, num);
          for (int k = 0; k < v.size(); k++) m[v[k]]++;
        }
    printf("%d\n", ans);
  }
  return 0;
}
