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 l, a[1001], b[1001], c[1001], g[1001], m, v, y, ans;
map<int, int> f;
int main() {
  cin >> l;
  for (int i = 1; i <= l; i++) {
    scanf("%d", &a[i]);
    f[a[i]]++;
  }
  sort(a + 1, a + l + 1);
  for (int i = 1; i <= l; i++) {
    for (int j = 1; j <= l; j++) {
      if (i != j) {
        if (i > 1 && a[i - 1] == a[i]) break;
        if (a[i] == 0 && a[j] == 0) {
          ans = max(ans, f[0]);
          continue;
        }
        map<int, int> d;
        g[1] = a[i];
        g[2] = a[j];
        d[g[1]]++;
        d[g[2]]++;
        y = 2;
        while (d[g[y] + g[y - 1]] < f[g[y] + g[y - 1]]) {
          y++;
          g[y] = g[y - 1] + g[y - 2];
          d[g[y]]++;
        }
        ans = max(ans, y);
      }
    }
  }
  cout << ans;
}
