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, a[2001], ans, f[2001];
map<int, int> mp;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    mp[a[i]]++;
  }
  sort(a + 1, a + n + 1);
  ans = max(ans, mp[0]);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (j > 1 && a[j] == a[j - 1]) continue;
      if (i != j) {
        int f1, f2, sum, w;
        if (!a[i] && !a[j]) {
          continue;
        }
        map<int, int> mp2;
        sum = 2;
        mp2[a[i]]++;
        mp2[a[j]]++;
        f1 = a[i];
        f2 = a[j];
        while (++mp2[w = f1 + f2] <= mp[f1 + f2]) {
          sum++;
          f1 = f2;
          f2 = w;
        }
        ans = max(ans, sum);
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}
