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 int a[1001];
map<long long int, int> used;
int ans = 0;
map<pair<long long int, long long int>, bool> us;
void rec(long long int x, long long int y, int kol) {
  ans = max(ans, kol);
  long long int z = x + y;
  if (z > 1000000000 || z < -1000000000) return;
  x = y;
  y = z;
  if (used[z] > 0) {
    used[z]--;
    rec(x, y, kol + 1);
    used[z]++;
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> a[i];
    used[a[i]]++;
  }
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (us[make_pair(a[i], a[j])] || i == j) continue;
      us[make_pair(a[i], a[j])] = true;
      used[a[i]]--;
      used[a[j]]--;
      rec(a[i], a[j], 2);
      used[a[i]]++;
      used[a[j]]++;
    }
  }
  cout << ans << endl;
  return 0;
}
