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 = 1e3 + 50;
int n, a[maxn], zero;
map<int, int> ha;
int dfs(int x, int y) {
  if (!ha.count(x + y) || ha[x + y] == 0) return 0;
  ha[x + y]--;
  int res = dfs(y, x + y);
  ha[x + y]++;
  return res + 1;
}
int main(int argc, char *argv[]) {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
    ha[a[i]]++;
    if (a[i] == 0) zero++;
  }
  sort(a + 1, a + n + 1);
  int ans = 0;
  for (int i = 1; i <= n; ++i)
    for (int j = 1; j <= n; ++j) {
      if (i != j) {
        if (a[i] == 0 && a[j] == 0)
          ans = max(ans, zero);
        else {
          ha[a[i]]--, ha[a[j]]--;
          ans = max(ans, dfs(a[i], a[j]) + 2);
          ha[a[i]]++, ha[a[j]]++;
        }
      }
    }
  printf("%d\n", ans);
  return 0;
}
