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>
int n, a[1005], c[1005], ans = 0;
std::map<int, int> b;
std::queue<int> q;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    ++b[a[i]];
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (i != j) {
        if (a[i] == 0 && a[j] == 0) {
          if (b[0] > ans) ans = b[0];
          continue;
        }
        int p = 2;
        c[1] = a[i];
        c[2] = a[j];
        --b[c[1]], --b[c[2]];
        q.push(c[1]), q.push(c[2]);
        for (int k = 3; k <= n; k++) {
          c[k] = c[k - 1] + c[k - 2];
          if (b[c[k]]) {
            b[c[k]]--;
            q.push(c[k]);
            ++p;
          } else
            break;
        }
        if (p > ans) ans = p;
        if (ans == n) break;
        while (!q.empty()) ++b[q.front()], q.pop();
      }
  printf("%d\n", ans);
  return 0;
}
