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;
const int inf = 1e9;
int n, a[N];
int f[N];
map<int, int> tal;
int main() {
  scanf("%d", &n);
  int cnt = 0;
  for (int i = (1), _x = (n); i <= _x; i++)
    scanf("%d", &a[i]), cnt += (a[i] == 0), tal[a[i]]++;
  for (int i = (1), _x = (n); i <= _x; i++) {
    for (int j = (1), _x = (n); j <= _x; j++) {
      if (i == j) continue;
      if (a[i] == 0 && a[j] == 0) continue;
      f[0] = a[i], f[1] = a[j];
      tal[a[i]]--, tal[a[j]]--;
      int tot = 0;
      for (int k = 2;; k++) {
        f[k] = f[k - 1] + f[k - 2];
        if (tal[f[k]] > 0)
          tal[f[k]]--;
        else {
          tot = k - 1;
          break;
        }
      }
      cnt = max(cnt, tot + 1);
      for (int k = (0), _x = (tot); k <= _x; k++) tal[f[k]]++;
    }
  }
  printf("%d\n", cnt);
  return 0;
}
