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[1005];
unordered_map<long long, int> mp;
long long t[105], cnt;
int main() {
  mp.max_load_factor(0.1);
  mp.reserve(262144);
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%I64d", &a[i]);
  for (int i = 0; i < n; i++) mp[a[i]]++;
  int res = 2;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      if (a[i] == 0 && a[j] == 0)
        res = max(res, mp[0]);
      else {
        t[0] = a[i];
        t[1] = a[j];
        long long now = t[0] + t[1];
        cnt = 2;
        while (abs(now) <= 1000000000) {
          t[cnt++] = now;
          now = t[cnt - 2] + t[cnt - 1];
        }
        t[cnt++] = now;
        mp[t[0]]--;
        mp[t[1]]--;
        for (int k = 2; k < cnt; k++) {
          if (mp.find(t[k]) == mp.end() || mp[t[k]] == 0) {
            res = max(res, k);
            for (int tt = 0; tt < k; tt++) mp[t[tt]]++;
            break;
          } else
            mp[t[k]]--;
        }
      }
    }
  printf("%d\n", res);
  return 0;
}
