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;
map<int, int> cnt;
const int N = 1005;
vector<int> p;
int st[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    int a;
    scanf("%d", &a);
    cnt[a]++;
    p.push_back(a);
  }
  sort(p.begin(), p.end());
  p.resize(distance(p.begin(), unique(p.begin(), p.end())));
  int ans = 2;
  for (int i = 0; i < p.size(); i++) {
    for (int j = 0; j < p.size(); j++) {
      cnt[p[i]]--;
      if (cnt[p[j]] == 0) {
        cnt[p[i]]++;
        continue;
      }
      cnt[p[j]]--;
      int a = p[i], b = p[j], add = 2, sz = 0;
      while (cnt[a + b]) {
        cnt[a + b]--;
        st[sz++] = a + b;
        b = b + a;
        a = b - a;
        add++;
      }
      ans = max(ans, add);
      cnt[p[i]]++;
      cnt[p[j]]++;
      for (int i = 0; i < sz; i++) cnt[st[i]]++;
    }
  }
  cout << ans << endl;
  return 0;
}
