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;
const int maxn = 1010;
map<int, int> cnt, tmp_cnt;
vector<int> used;
int a[maxn];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", a + i);
    cnt[a[i]]++;
    tmp_cnt[a[i]]++;
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      for (int k = 0; k < (int)used.size(); k++)
        tmp_cnt[used[k]] = cnt[used[k]];
      used.clear();
      int last = a[i], last_sec = a[j];
      tmp_cnt[last]--;
      tmp_cnt[last_sec]--;
      used.push_back(last);
      used.push_back(last_sec);
      int len = 2;
      while (tmp_cnt[last + last_sec]) {
        if (last == 0 && last_sec == 0) {
          len += tmp_cnt[0];
          break;
        }
        int next_last = last + last_sec;
        tmp_cnt[next_last]--;
        last_sec = last;
        last = next_last;
        len++;
        used.push_back(last);
      }
      ans = max(ans, len);
    }
  }
  printf("%d\n", ans);
  return 0;
}
