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;
map<int, int> mp;
int a[1001], ans;
int cnt;
int t;
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    mp[a[i]]++;
  }
  sort(a + 1, a + 1 + n);
  ans = 2;
  ans = max(mp[0], ans);
  int a1, a2;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      if ((long long)i == j) continue;
      if (a[i] == 0 && a[j] == 0) continue;
      if (i > 1 && a[i] == a[i - 1]) break;
      map<int, int> mp2;
      mp2[a[i]]++, mp2[a[j]]++;
      cnt = 2;
      a1 = a[i], a2 = a[j];
      while (++mp2[a1 + a2] <= mp[a1 + a2]) {
        cnt++, t = a1;
        a1 = a2, a2 += t;
      }
      ans = max(ans, cnt);
    }
  printf("%d", ans);
  return 0;
}
