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 = 1010;
unordered_map<int, int> id;
int a[maxn], cnt[maxn];
int cur[maxn], used[maxn];
int it;
bool u[maxn][maxn];
int main() {
  int n;
  cin >> n;
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (!id[a[i]]) id[a[i]] = i;
    cnt[id[a[i]]]++;
    if (!a[i]) ans++;
  }
  int res, x, y, tmp;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      if (a[i] == 0 && a[j] == 0) continue;
      if (u[id[a[i]]][id[a[j]]]) continue;
      u[id[a[i]]][id[a[j]]] = 1;
      it++;
      cur[id[a[i]]] = cur[id[a[j]]] = 0;
      used[id[a[i]]] = used[id[a[j]]] = it;
      cur[id[a[i]]]++;
      cur[id[a[j]]]++;
      x = a[i], y = a[j];
      res = 2;
      while (1) {
        if (!id[x + y] ||
            (used[id[x + y]] == it && cur[id[x + y]] == cnt[id[x + y]]))
          break;
        res++;
        if (used[id[x + y]] != it) {
          used[id[x + y]] = it;
          cur[id[x + y]] = 0;
        }
        cur[id[x + y]]++;
        tmp = y;
        y += x;
        x = tmp;
      }
      ans = max(ans, res);
    }
  }
  cout << ans;
  return 0;
}
