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;
void debug(vector<long long> v) {
  for (int i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << "\n";
}
long long arr[1010];
map<long long, long long> imap, cur;
int main() {
  long long t, z, i, j, k, n, m, p, q, r, s, ans;
  scanf("%lld", &n);
  for (i = 0; i < n; i++) {
    scanf("%lld", &arr[i]);
    imap[arr[i]]++;
  }
  long long a, b, c;
  long long mx = 1e9 + 10;
  ans = 0;
  vector<long long> x;
  for (auto p : imap) x.push_back(p.first);
  n = x.size();
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      cur.clear();
      a = x[i];
      b = x[j];
      cur[a]++;
      cur[b]++;
      if (cur[a] > imap[a] || cur[b] > imap[b]) continue;
      long long cnt = 2;
      while (abs(b) <= mx) {
        c = b;
        b = a + b;
        a = c;
        cur[b]++;
        if (imap.find(b) == imap.end() || imap[b] < cur[b]) break;
        cnt++;
      }
      ans = max(ans, cnt);
    }
  }
  printf("%lld\n", ans);
  return 0;
}
