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 Max(int &x, int y) {
  if (x < y) x = y;
}
int main() {
  int n;
  scanf("%d", &n);
  vector<int> v(n);
  unordered_map<int, int> st;
  for (auto &e : v) scanf("%d", &e), ++st[e];
  int ans = max(2, (int)st[0]);
  auto solve = [&](int x, int y) {
    vector<int> ret({x, y});
    for (auto e : ret) --st[e];
    for (;;) {
      int nxt = ret.back() + ret[ret.size() - 2];
      if (st.count(nxt) == 0 || st[nxt] == 0) break;
      ret.push_back(nxt);
      --st[nxt];
    }
    Max(ans, (int)ret.size());
    for (auto e : ret) ++st[e];
    ;
  };
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j)
      if (i != j && (v[i] != 0 || v[j] != 0)) solve(v[i], v[j]);
  printf("%d\n", ans);
}
