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 a[1010];
int main() {
  ios::sync_with_stdio(false);
  map<long long, int> x;
  long long n;
  cin >> n;
  long long totzero = 0;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    x[a[i]]++;
    if (a[i] == 0) totzero++;
  }
  long long mm = 0;
  map<pair<long long, long long>, bool> qq;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i != j && !qq[make_pair(a[i], a[j])]) {
        qq[make_pair(a[i], a[j])] = 1;
        map<long long, int> used;
        if (a[i] == 0 && a[j] == 0) {
          mm = max(mm, totzero);
          continue;
        }
        long long t = a[i] + a[j];
        used[a[i]]++;
        used[a[j]]++;
        int v = a[j];
        long long r = 2;
        while (x[t] != 0 && used[t] < x[t]) {
          used[t]++;
          int w = t;
          t += v;
          v = w;
          r++;
        }
        mm = max(mm, r);
      }
    }
  }
  cout << mm;
}
