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[1001], ans, i, j, x;
map<int, int> m;
map<pair<int, int>, bool> used;
vector<int> v;
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    m[a[i]]++;
  }
  for (i = 1; i <= n; i++) {
    for (j = 1; j <= n; j++) {
      if (i == j) continue;
      if (used[make_pair(a[i], a[j])]) continue;
      used[make_pair(a[i], a[j])] = true;
      m[a[i]]--;
      m[a[j]]--;
      v.push_back(a[i]);
      v.push_back(a[j]);
      x = a[i] + a[j];
      while (m[x]) {
        m[x]--;
        v.push_back(x);
        x += v[v.size() - 2];
      }
      ans = max(ans, (int)v.size());
      while (!v.empty()) {
        m[v.back()]++;
        v.pop_back();
      }
    }
  }
  cout << ans << endl;
}
