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 = 1005;
int n, a[N], vis[N];
map<int, int> dem, dem_now;
int calc(int i, int j) {
  int ret = 2;
  vector<int> val;
  int ni = a[i], nj = a[j];
  dem_now[ni] = 1;
  dem_now[nj] = 1;
  val.push_back(ni);
  val.push_back(nj);
  while (true) {
    int need = ni + nj;
    if (dem_now[need] < dem[need]) {
      val.push_back(need);
      dem_now[need]++;
      ni = nj;
      nj = need;
      ret++;
    } else
      break;
  }
  for (auto& x : val) dem_now[x] = 0;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    dem[a[i]]++;
  }
  int res = dem[0];
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      if (a[i] == 0 && a[j] == 0) continue;
      res = max(res, calc(i, j));
    }
  }
  cout << res;
}
