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 dx8[] = {0, 0, 1, -1, 1, -1, 1, -1};
int dy8[] = {-1, 1, 0, 0, -1, -1, 1, 1};
int dy[] = {1, -1, 0, 0};
int dx[] = {0, 0, 1, -1};
int X[6] = {0, 0, 0, -1, 0, 1};
int Y[6] = {0, 0, -1, 0, 1, 0};
int Z[6] = {1, -1, 0, 0, 0, 0};
int main() {
  int n;
  cin >> n;
  vector<long long> v(n);
  unordered_map<long long, long long> mp, mmp;
  for (int i = 0; i < n; i++) {
    cin >> v[i];
    mp[v[i]]++;
  }
  long long mx = -1;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if ((!v[i] && !v[j]) || i == j) continue;
      vector<long long> vec;
      long long c = 2;
      long long x = v[i] + v[j], last = v[j];
      vec.push_back(v[i]);
      vec.push_back(v[j]);
      mp[v[i]]--;
      mp[v[j]]--;
      while (mp[x]) {
        mp[x]--;
        vec.push_back(x);
        c++;
        long long y = x;
        x += last;
        last = y;
      }
      for (int k = 0; k < vec.size(); k++) mp[vec[k]]++;
      mx = max(mx, c);
    }
  }
  cout << max(mx, mp[0]) << "\n";
}
