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 n;
int a[1010];
int x[2];
int answ, d;
map<int, int> md;
int main(void) {
  scanf("%d", &n);
  for (register int i = 0; i < n; ++i) scanf("%d", a + i), ++md[a[i]];
  for (register int i = 0; i < n; ++i)
    for (register int j = 0; j < n; ++j) {
      if (i == j) continue;
      x[0] = a[i];
      x[1] = a[j];
      int next = x[0] + x[1];
      d = 2;
      if (!x[0] && !x[1]) {
        answ = max(answ, md[0]);
        continue;
      }
      --md[x[0]], --md[x[1]];
      while (md.find(next) != md.end() && md[next]) {
        --md[next], ++d;
        x[0] = next;
        swap(x[1], x[0]);
        next = x[0] + x[1];
      }
      x[0] = a[i];
      x[1] = a[j];
      ++md[x[0]], ++md[x[1]];
      next = x[0] + x[1];
      answ = max(answ, d);
      while (d > 2) {
        ++md[next], --d;
        x[0] = next;
        swap(x[1], x[0]);
        next = x[0] + x[1];
      }
    }
  printf("%d\n", answ);
  return 0;
}
