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;
struct _ {
  ios_base::Init i;
  _() { ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL); }
} _;
vector<long long> a;
long long n, pp, p, curr, len, max_len;
unordered_map<long long, long long> vis;
unordered_map<long long, long long> vis2;
stack<long long> st;
int main() {
  cin >> n;
  a.resize(n, 0);
  for (int i = 0; i <= n - 1; i++) cin >> a[i], vis[a[i]]++;
  for (int i = 1; i <= n - 1; i++) {
    for (int j = 0; j <= i - 1; j++) {
      if (a[i] == 0 and a[j] == 0) {
        max_len = max(max_len, vis[0]);
        continue;
      }
      pp = a[j], p = a[i], curr = a[i] + a[j];
      vis[pp]--;
      st.push(pp);
      vis[p]--;
      st.push(p);
      len = 2;
      while (vis[curr] > 0) {
        vis[curr]--;
        st.push(curr);
        pp = p, p = curr;
        curr = pp + p;
        len++;
      }
      max_len = max(max_len, len);
      while (!st.empty()) vis[st.top()]++, st.pop();
      pp = a[i], p = a[j], curr = a[i] + a[j];
      vis[pp]--;
      st.push(pp);
      vis[p]--;
      st.push(p);
      len = 2;
      while (vis[curr] > 0) {
        vis[curr]--;
        st.push(curr);
        pp = p, p = curr;
        curr = pp + p;
        len++;
      }
      max_len = max(max_len, len);
      while (!st.empty()) vis[st.top()]++, st.pop();
    }
  }
  cout << max_len;
  return 0;
}
