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<int, int> mp;
int ans = 0;
void check(int i, int j, int len = 2) {
  int nx = i + j;
  ans = max(ans, len);
  if (mp[nx] > 0) {
    mp[nx]--;
    check(nx, i, len + 1);
    mp[nx]++;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  vector<int> ar;
  for (int i = 0; i < n; i++) {
    int el;
    cin >> el;
    mp[el]++;
    ar.push_back(el);
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      if (ar[i] == ar[j] && ar[i] == 0) {
        ans = max(ans, mp[0]);
        continue;
      }
      mp[ar[i]]--;
      mp[ar[j]]--;
      check(ar[i], ar[j]);
      mp[ar[i]]++;
      mp[ar[j]]++;
    }
  }
  cout << ans << endl;
  return 0;
}
