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>
const int md = 1e9 + 7;
const long long hs = 199;
const long double eps = 1e-7, pi = acos(-1);
using namespace std;
int n, mx = 2;
vector<int> a;
unordered_map<int, int> m;
queue<int> q;
int f(int k) { return k % 3; }
int check(int x, int y) {
  int cnt = 2, k[3];
  m[x]--, m[y]--;
  q.push(x), q.push(y);
  k[0] = x, k[1] = y;
  while (m[k[f(cnt + 1)] + k[f(cnt + 2)]]) {
    m[k[f(cnt + 1)] + k[f(cnt + 2)]]--;
    k[f(cnt)] = k[f(cnt + 1)] + k[f(cnt + 2)];
    q.push(k[f(cnt)]);
    cnt++;
  }
  while (!q.empty()) m[q.front()]++, q.pop();
  return cnt;
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  a.resize(n);
  for (int i = 0; i < n; i++) cin >> a[i], m[a[i]]++;
  mx = max(mx, m[0]);
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j < n; j++)
      if (a[i] != 0 || a[j] != 0)
        mx = max(max(mx, check(a[i], a[j])), check(a[j], a[i]));
  cout << mx << '\n';
}
