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>
const int SZ = 1e3 + 5;
using namespace std;
int n;
int a[SZ];
map<int, int> mp;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", a + i), mp[a[i]]++;
  int res = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i != j && !(!a[i] && !a[j])) {
        vector<int> tmp;
        int x = a[i], y = a[j];
        mp[x]--, mp[y]--;
        tmp.push_back(x), tmp.push_back(y);
        while (mp[x + y] > 0) {
          tmp.push_back(x + y);
          mp[x + y]--;
          int z = x + y;
          x = y;
          y = z;
        }
        for (int k = 0; k < (int)tmp.size(); k++) mp[tmp[k]]++;
        res = max(res, (int)tmp.size());
      }
  cout << max(res, mp[0]);
}
