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 N = 1002;
map<int, int> cnt;
set<pair<int, int> > st;
int n, res, a[N];
void init() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    cnt[a[i]]++;
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      map<int, int> ct;
      ct.clear();
      int u = a[i], v = a[j], w = u + v, l = 2;
      if (st.count(make_pair(u, v)))
        continue;
      else
        st.insert(make_pair(u, v));
      ct[u]++, ct[v]++;
      while (1) {
        if (u == 0 && v == 0) {
          l += cnt[0] - ct[0];
          break;
        }
        ct[w]++;
        if (ct[w] > cnt[w]) break;
        l++;
        u = v, v = w, w = u + v;
      }
      res = max(res, l);
    }
  }
  printf("%d\n", res);
}
void fread() {
  freopen("input.txt", "r", stdin);
  freopen("output.txt", "w", stdout);
}
int main() {
  init();
  return 0;
}
