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;
long long n, a[1005];
map<long long, int> x;
int ans;
void f(long long a, long long b) {
  long long cur = a + b;
  vector<long long> q;
  q.push_back(a);
  q.push_back(b);
  int len = 2;
  x[a]--;
  x[b]--;
  while (x[cur] > 0) {
    x[cur]--;
    q.push_back(cur);
    a = b;
    b = cur;
    cur = a + b;
    len++;
  }
  ans = max(len, ans);
  for (int i = 0; i < q.size(); i++) x[q[i]]++;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    x[a[i]]++;
  }
  ans = x[0];
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (a[i] == a[j] && a[i] == 0) continue;
      f(a[i], a[j]);
      f(a[j], a[i]);
    }
  }
  cout << ans;
}
