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 arr[1111];
unordered_map<int, int> mp;
int main() {
  int n;
  cin >> n;
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
    ++mp[arr[i]];
  }
  set<pair<int, int> > st;
  int ans = 0;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      if (i == j) continue;
      pair<int, int> p = make_pair(arr[i], arr[j]);
      int v = 2;
      if (st.count(p) == 0) {
        st.insert(p);
        long long a = arr[i];
        long long b = arr[j];
        --mp[arr[i]];
        --mp[arr[j]];
        vector<int> vt;
        while (v < n) {
          long long t = a + b;
          if (mp[t] == 0) break;
          vt.push_back(t);
          --mp[t];
          ++v;
          a = b;
          b = t;
        }
        for (int i = 0; i < vt.size(); ++i) ++mp[vt[i]];
        ++mp[arr[i]];
        ++mp[arr[j]];
      }
      ans = max(ans, v);
    }
  }
  cout << ans << endl;
  return 0;
}
