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;
void init_ios() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
}
const int N = 1010;
int n, a[N], res[N][N];
bool vis[N][N];
int wyn, ile[N], t = 1;
map<int, int> m;
void jebaj(int x, int y, int z) {
  wyn = max(wyn, z);
  if (ile[m[x + y]] > 0) {
    --ile[m[x + y]];
    jebaj(y, x + y, z + 1);
    ++ile[m[x + y]];
  }
}
int main() {
  init_ios();
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    if (m[a[i]] == 0) m[a[i]] = t++;
    ile[m[a[i]]]++;
  }
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j)
      if (i != j && !vis[m[a[i]]][m[a[j]]]) {
        vis[m[a[i]]][m[a[j]]] = true;
        --ile[m[a[i]]];
        --ile[m[a[j]]];
        jebaj(a[i], a[j], 2);
        ++ile[m[a[i]]];
        ++ile[m[a[j]]];
      }
  cout << wyn << "\n";
}
