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[1005], used[1005];
map<int, int> vis;
inline int dfs(int i, int j) {
  vis[j]--;
  int ret = 0;
  if (vis[i + j]) {
    ret = dfs(j, i + j) + 1;
  }
  vis[j]++;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  int zeroes = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    vis[a[i]]++;
  }
  int ret = max(0, vis[0] - 1);
  for (int i = 0; i < n; i++) {
    vis[a[i]]--;
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      int index = 0;
      int a1 = a[i], b1 = a[j];
      if (a1 == 0 && b1 == 0) continue;
      int temp = 0;
      vis[b1]--;
      index++;
      used[0] = b1;
      while (vis[a1 + b1]) {
        vis[a1 + b1]--;
        used[index] = a1 + b1;
        index++;
        temp = a1 + b1;
        a1 = b1;
        b1 = temp;
      }
      ret = max(ret, index);
      index--;
      while (index >= 0) {
        vis[used[index]]++;
        index--;
      }
    }
    vis[a[i]]++;
  }
  cout << ret + 1;
}
