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 main() {
  int n;
  scanf("%d", &n);
  int s[n];
  map<int, int> ct;
  for (int i = 0; i < int(n); i++) {
    scanf("%d", &s[i]);
    ct[s[i]]++;
  }
  int ans = 2;
  long x, f1, f2, f3;
  pair<int, int> aux2;
  for (int i = 0; i < int(n); i++)
    for (int j = 0; j < int(n); j++) {
      if (i == j) continue;
      if (s[i] == 0 && s[i] == s[j]) {
        ans = max(ans, ct[0]);
        continue;
      }
      x = 2;
      vector<pair<int, int> > aux;
      aux2.first = s[i];
      aux2.second = ct[s[i]];
      ct[s[i]]--;
      aux.push_back(aux2);
      aux2.first = s[j];
      aux2.second = ct[s[j]];
      aux.push_back(aux2);
      ct[s[j]]--;
      f1 = s[j];
      f2 = s[j] + s[i];
      while (ct[f2] > 0) {
        aux2.first = f2;
        aux2.second = ct[f2];
        aux.push_back(aux2);
        x++;
        ct[f2]--;
        f3 = f1;
        f1 = f2;
        f2 += f3;
      }
      for (pair<int, int> p : aux) {
        if (p.second > ct[p.first]) {
          ct[p.first] = p.second;
        }
      }
      if (x > ans) ans = x;
    }
  printf("%d", ans);
  return 0;
}
