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;
map<long long, int> M;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int N, i;
  cin >> N;
  vector<int> A(N + 1);
  for (i = 0; i < N; ++i) {
    cin >> A[i];
    M[A[i]]++;
  }
  int ans = 0;
  for (i = 0; i < N; ++i)
    for (int j = 0; j < N; ++j) {
      if (A[i] == A[j] && A[i] == 0) continue;
      if (i == j) continue;
      int temp = 2;
      long long f = A[i], s = A[j];
      M[f]--;
      M[s]--;
      while (M.find(f + s) != M.end()) {
        if (!M[f + s]) break;
        M[f + s]--;
        swap(f, s);
        s += f;
        ++temp;
      }
      ans = max(ans, temp);
      f = A[i], s = A[j];
      M[f]++;
      M[s]++;
      temp -= 2;
      while (temp) {
        M[f + s]++;
        swap(f, s);
        s += f;
        --temp;
      }
    }
  ans = max(ans, M[0]);
  cout << ans;
  return 0;
}
