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;
int inf = 1e9;
map<int, short> p, u;
short n;
int a[1005], b[1005];
int main() {
  scanf("%hd", &n);
  for (short i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    ++p[a[i]];
  }
  short ans = max(short(2), p[0]);
  for (short i = 0; i < n; ++i)
    for (short j = 0; j < n; ++j) {
      if (i == j || (!a[i] && !a[j])) continue;
      ++u[a[i]];
      ++u[a[j]];
      b[0] = a[i];
      b[1] = a[j];
      short cur = 2;
      for (; b[cur - 1] + b[cur - 2] <= inf; ++cur) {
        b[cur] = b[cur - 1] + b[cur - 2];
        if (u[b[cur]] + 1 > p[b[cur]]) break;
        ++u[b[cur]];
        ans = max(ans, short(cur + 1));
      }
      for (int x = 0; x <= cur; ++x) u[b[x]] = 0;
    }
  printf("%hd", ans);
  return 0;
}
