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[1111], b[1111];
map<int, int> arr;
map<pair<int, int>, bool> use;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    arr[a[i]]++;
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j || use[{a[i], a[j]}]) continue;
      use[{a[i], a[j]}] = true;
      int k = 2;
      b[0] = a[i];
      b[1] = a[j];
      arr[a[i]]--;
      arr[a[j]]--;
      while (arr[b[k - 1] + b[k - 2]] > 0) {
        b[k] = b[k - 1] + b[k - 2];
        arr[b[k]]--;
        k++;
      }
      if (k > ans) ans = k;
      for (int j = 0; j < k; j++) arr[b[j]]++;
    }
  }
  cout << ans << endl;
}
