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>
#pragma comment(linker, "/STACK:500000000")
using namespace std;
int a[1009];
int main() {
  int n, i, j;
  scanf("%d", &n);
  set<pair<long long, long long> > cache;
  multiset<long long> s;
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    s.insert(a[i]);
  }
  int res = 0;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      if (i == j) continue;
      long long A = a[i], B = a[j], C;
      if (cache.count({A, B})) continue;
      cache.insert({A, B});
      s.erase(s.find(A));
      s.erase(s.find(B));
      vector<long long> bu = {A, B};
      int len = 2;
      do {
        C = A + B;
        if (!s.count(C)) break;
        bu.push_back(C);
        s.erase(s.find(C));
        len++;
        A = B;
        B = C;
      } while (abs(C) < 1e11);
      res = max(res, len);
      for (auto e : bu) s.insert(e);
    }
  }
  printf("%d\n", res);
}
