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;
const int N = 1e3 + 10, Sq = 324;
map<int, int> m;
int a[N], s[N];
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, ans = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    m[a[i]]++;
  }
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      if (a[i] == 0 && a[j] == 0) {
        ans = max(ans, m[0]);
        continue;
      }
      int cnt = 2;
      s[1] = a[i];
      s[2] = a[j];
      m[s[1]]--;
      m[s[2]]--;
      while (m[s[cnt] + s[cnt - 1]]) {
        s[++cnt] = s[cnt - 1] + s[cnt - 2];
        m[s[cnt]]--;
      }
      ans = max(ans, cnt);
      for (int k = 1; k <= cnt; k++) m[s[k]]++;
    }
  cout << ans;
  return 0;
}
