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;
inline void pisz(int n) { printf("%d\n", n); }
template <typename T, typename TT>
ostream& operator<<(ostream& s, pair<T, TT> t) {
  return s << "(" << t.first << "," << t.second << ")";
}
template <typename T>
ostream& operator<<(ostream& s, vector<T> t) {
  for (int(i) = 0; (i) < (((int)((t).size()))); ++(i)) s << t[i] << " ";
  return s;
}
int main() {
  int(n);
  scanf("%d", &(n));
  set<int> distinct;
  unordered_map<int, int> m;
  for (int(i) = 0; (i) < (n); ++(i)) {
    int(x);
    scanf("%d", &(x));
    m[x]++;
    distinct.insert(x);
  }
  int ma = 2;
  for (int a0 : distinct)
    for (int b0 : distinct) {
      if (a0 == b0 && m[a0] < 2) continue;
      int len = 2;
      int a = a0, b = b0;
      --m[a];
      --m[b];
      while (1) {
        int c = a + b;
        unordered_map<int, int>::iterator it = m.find(c);
        if (it == m.end() || it->second < 1) break;
        --it->second;
        ++len;
        a = b;
        b = c;
      }
      (ma) = max((ma), (len));
      ;
      a = a0;
      b = b0;
      ++m[a];
      ++m[b];
      for (int(k) = (3); (k) <= (len); ++k) {
        int c = a + b;
        m[c]++;
        a = b;
        b = c;
      }
    }
  pisz(ma);
}
