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 long long maxn = 1005;
long long n;
long long a[maxn], mx = 0;
unordered_map<long long, long long> mp;
void go(long long a, long long b) {
  if (!a && !b) {
    mx = max(mx, mp[0]);
    return;
  }
  vector<long long> v;
  long long f1 = 0, f2 = 1, f3 = 1;
  long long res = 2;
  mp[a]--, mp[b]--;
  for (long long i = 1; i <= 100; i++) {
    long long t = a * f2 + b * f3;
    if (mp.count(t) && mp[t])
      res++, mp[t]--, v.push_back(t);
    else
      break;
    f1 = f2, f2 = f3, f3 = f1 + f2;
  }
  for (long long x : v) mp[x]++;
  mp[a]++, mp[b]++;
  mx = max(res, mx);
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> a[i];
    mp[a[i]]++;
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= n; j++) {
      if (i - j) go(a[i], a[j]);
    }
  }
  cout << mx;
}
