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 long long M = 1e9 + 7;
const long long INF = (1LL << 62);
inline long long pwr(long long base, long long n, long long m) {
  long long ans = 1;
  while (n > 0) {
    if (n & 1) ans = (ans * base) % m;
    base = (base * base) % m;
    n >>= 1;
  }
  return ans;
}
const int N = 1005;
int n;
long long a[N];
unordered_map<long long, int> f;
deque<long long> undo;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n;
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    f[a[i]]++;
  }
  if (n == 2) {
    cout << 2;
    return 0;
  }
  int ans = 2;
  for (long long i = 0; i < n; i++) {
    for (long long j = 0; j < n; j++) {
      if (i == j) continue;
      f[a[i]]--;
      f[a[j]]--;
      undo.push_back(a[i]);
      undo.push_back(a[j]);
      int cnt = 2;
      if (a[i] == 0 && a[j] == 0) {
        while (!undo.empty()) {
          f[undo[0]]++;
          undo.pop_front();
        }
        ans = max(ans, f[0]);
        continue;
      }
      long long prev = a[j], nxt = a[i] + a[j];
      while (f[nxt]) {
        cnt++;
        f[nxt]--;
        undo.push_back(nxt);
        long long temp = prev;
        prev = nxt;
        nxt += temp;
      }
      ans = max(ans, cnt);
      while (!undo.empty()) {
        f[undo[0]]++;
        undo.pop_front();
      }
    }
  }
  cout << ans;
  return 0;
}
