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;
const int mod = 1000000007;
long long i, j, t;
bool flag = false;
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t = 1;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> v(n);
    unordered_map<long long, long long> mp1;
    for (i = 0; i < n; i++) {
      cin >> v[i];
      mp1[v[i]]++;
    }
    if (mp1[0] > n / 2) {
      cout << mp1[0] << endl;
      continue;
    }
    long long len = 0;
    for (i = 0; i < n; i++) {
      for (j = 0; j < n; j++) {
        if (i != j) {
          if (v[i] == 0 && v[j] == 0)
            len = max(len, mp1[0]);
          else {
            vector<long long> temp;
            long long a = v[i], b = v[j], next = a + b;
            temp.push_back(a);
            temp.push_back(b);
            mp1[a]--;
            mp1[b]--;
            while (mp1.find(next) != mp1.end() && mp1[next] > 0) {
              temp.push_back(next);
              mp1[next]--;
              a = b;
              b = next;
              next = a + b;
            }
            long long tsize = temp.size();
            len = max(len, tsize);
            for (auto i : temp) mp1[i]++;
          }
        }
      }
    }
    cout << len << endl;
  }
}
