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 nmax = 1000 * 1000;
const int inf = 1000 * 1000 * 1000;
const long long infl = 1000ll * 1000ll * 1000ll * 1000ll * 1000ll * 1000ll;
const int mod = 1000 * 1000 * 1000 + 7;
const long double pi = acos(-1.0);
template <typename T>
ostream& operator<<(ostream& cout, const vector<T>& a) {
  if ((int)a.size() == 0) {
    return (cout << "()");
  }
  cout << "(" << a[0];
  for (int i = 1; i < (int)a.size(); i++) {
    cout << "; " << a[i];
  }
  return (cout << ")");
}
template <typename T1, typename T2>
ostream& operator<<(ostream& cout, const pair<T1, T2>& a) {
  return cout << "(" << a.first << "; " << a.second << ")";
}
int n, a[nmax];
map<long long, int> now;
int answer = 0;
long long f[nmax];
int main() {
  int cnt = 0;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    if (a[i] == 0) {
      answer++;
    }
    now[a[i]]++;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j || (a[j] == 0 && a[i] == 0)) {
        continue;
      }
      f[0] = a[i];
      f[1] = a[j];
      now[a[i]]--, now[a[j]]--;
      int sz = 2;
      while (sz < n) {
        f[sz] = f[sz - 1] + f[sz - 2];
        if (now[f[sz]]) {
          now[f[sz++]]--;
        } else {
          break;
        }
      }
      for (int q = 0; q < sz; q++) {
        now[f[q]]++;
      }
      answer = max(answer, sz);
    }
  }
  cout << max(cnt, answer) << endl;
}
