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 + 2;
int n;
int a[N];
map<int, int> m;
map<pair<int, int>, bool> checked;
int main() {
  cin >> n;
  int zero = 0;
  for (int i = 0; i < n; ++i) {
    scanf("%d", a + i);
    if (!a[i]) zero++;
    ;
    m[a[i]]++;
  }
  sort(a, a + n);
  int ans = -1;
  int help1, help2, help3;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j)
      if (i != j) {
        if (!a[i] and (!a[j])) continue;
        if (checked[make_pair(a[i], a[j])]) continue;
        checked[make_pair(a[i], a[j])] = true;
        map<int, int> m1;
        help3 = 2;
        m1[a[i]]++, m1[a[j]]++;
        help1 = a[i];
        help2 = a[j];
        while (m1[help1 + help2] < m[help1 + help2]) {
          m1[help1 + help2]++;
          help3++;
          help2 += help1;
          help1 = help2 - help1;
        }
        ans = max(ans, help3);
      }
  ans = max(ans, zero);
  cout << ans;
}
