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;
int n, res = 0;
map<int, int> cnt;
int a[1001], f[1001];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    cnt[a[i]]++;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      if (a[i] == a[j] && a[i] == 0)
        res = max(res, cnt[0]);
      else {
        int x = a[i], y = a[j];
        f[1] = x;
        f[2] = y;
        cnt[x]--;
        cnt[y]--;
        for (int k = 3; i <= n; k++) {
          int z = f[k - 1] + f[k - 2];
          if (cnt[z] > 0)
            f[k] = z, cnt[z]--;
          else {
            k--;
            res = max(res, k);
            for (int p = 1; p <= k; p++) cnt[f[p]]++;
            break;
          }
        }
      }
    }
  cout << res;
}
