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 N = 1e3 + 100;
int arr[N];
int cnt[N];
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  map<long long, int> mp;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    mp[arr[i]]++;
  }
  int mx = 2;
  sort(arr, arr + n);
  mx = max(mp[0], mx);
  for (int i = 0; i < n; i++) {
    if (i && arr[i] == arr[i - 1]) continue;
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      int a1 = arr[i], a2 = arr[j];
      if (a1 == 0 && a2 == 0) continue;
      int cnt = 2;
      map<long long, int> mp2;
      mp2[a1]++, mp2[a2]++;
      while (++mp2[a1 + a2] <= mp[a1 + a2]) {
        int t = a1;
        cnt++, a1 = a2, a2 += t;
      }
      mx = max(cnt, mx);
    }
  }
  cout << mx << endl;
  return 0;
}
