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;
long long mod = 1000000007, oo = 1000000009;
void Emsawy() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
}
int dx[] = {1, -1, 0, 0, 1, -1, 1, -1};
int dy[] = {0, 0, 1, -1, 1, -1, -1, 1};
vector<vector<int> > adj;
const long long N = 1000000 + 6;
int n, m, k, w;
int in, M[N];
map<int, int> pq;
inline void extract(int x) {
  int cnt = 0;
  while (x > 0) {
    if (x & 1) pq[cnt]++;
    cnt++;
    x >>= 1;
  }
}
inline int pw(int x) { return (1 << x); }
void pr() {
  for (map<int, int>::iterator it = pq.begin(); it != pq.end(); it++)
    cout << it->first << " " << it->second << " " << endl;
}
int main() {
  int a, b, c, n;
  while (cin >> n) {
    vector<long long> v(n);
    map<long long, int> mp;
    for (int i = 0; i < n; i++) cin >> v[i], mp[v[i]]++;
    int si = v.size();
    sort(((v).begin()), ((v).end()));
    int res = 0;
    for (int i = 0; i < si; i++) {
      int l = v[i];
      mp[l]--;
      for (int j = 0; j < si; j++) {
        if (i == j) continue;
        int r = v[j], lo = 0;
        mp[r]--;
        int cl = l, cr = r;
        while (mp[l + r] > 0) {
          int t = l + r;
          l = r;
          r = t;
          lo++;
          mp[t]--;
        }
        while (cl != l || cr != r) {
          mp[r]++;
          int t = r - l;
          r = l;
          l = t;
        }
        mp[r]++;
        res = max(res, lo);
      }
      mp[l]++;
    }
    cout << res + 2 << endl;
  }
  return 0;
}
