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 MAXN = 1010;
long long v[MAXN];
unordered_map<long long, int> mp;
map<pair<long long, long long>, bool> mp2;
long long seq[1010];
int main() {
  ios::sync_with_stdio(false);
  ;
  int n;
  cin >> n;
  for (int i = 0; i < (n); ++i) {
    cin >> v[i];
    ++mp[v[i]];
  }
  int ans = 2;
  for (int i = 0; i < (n); ++i)
    for (int j = 0; j < (n); ++j)
      if (i != j) {
        long long a = v[i];
        long long b = v[j];
        if (mp2.count({a, b}))
          continue;
        else
          mp2[{a, b}] = true;
        int pans = 2;
        seq[0] = a;
        seq[1] = b;
        --mp[a];
        --mp[b];
        while (true) {
          long long next = a + b;
          auto it = mp.find(next);
          if (it != mp.end() && it->second > 0) {
            --mp[next];
            a = b;
            b = next;
            seq[pans++] = next;
          } else
            break;
        }
        ans = max(ans, pans);
        for (int k = 0; k < (pans); ++k) ++mp[seq[k]];
      }
  cout << ans << "\n";
  return 0;
}
