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 = 1013;
int n;
long long a[N];
map<pair<long long, long long>, int> s;
unordered_map<int, int> used;
unordered_map<int, int> cnt;
int dfs(long long a, long long b) {
  if (used[b] >= cnt[b]) {
    return 0;
  }
  used[b]++;
  int r = dfs(b, a + b);
  used[b]--;
  return r + 1;
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0);
  cin >> n;
  int zeros = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    zeros += (a[i] == 0);
    cnt[a[i]]++;
  }
  int res = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i != j && (a[i] || a[j])) {
        used[a[i]]++;
        res = max(res, 1 + dfs(a[i], a[j]));
        used[a[i]]--;
      }
  cout << max(zeros, res) << endl;
}
