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 GCC optimize(2)
using namespace std;
const int mod = 1e9 + 7;
const double eps = 1e-6;
const int N = 1e3 + 7;
int a[N], f[N];
map<int, int> mp;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i <= n - 1; i++) {
    scanf("%d", &a[i]);
    mp[a[i]]++;
  }
  int res = 0;
  for (int i = 0; i <= n - 1; i++)
    for (int j = 0; j <= n - 1; j++)
      if (i != j) {
        if (!a[i] && !a[j]) {
          res = max(res, mp[0]);
          continue;
        }
        f[1] = a[i], f[2] = a[j];
        int c = 2, f1 = a[i], f2 = a[j], w = f1 + f2;
        mp[f1]--;
        mp[f2]--;
        while (mp[w]) {
          mp[w]--;
          f[++c] = w;
          f1 = f2;
          f2 = w;
          w = f1 + f2;
        }
        res = max(res, c);
        for (int i = 1; i <= c; i++) mp[f[i]]++;
      }
  printf("%d\n", res);
  return 0;
}
