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<long long, int> mp;
long long a[1005], sum[1005];
int main() {
  int n;
  while (cin >> n) {
    long long ans = 2;
    long long maxx = -1e+9, res = 0;
    for (int i = 1; i <= n; i++) {
      cin >> a[i];
      if (maxx < a[i]) maxx = a[i];
      if (a[i] == 0) res++;
      mp[a[i]]++;
    }
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        vector<long long> v;
        if (i == j) continue;
        if (a[i] == 0 && a[j] == 0) continue;
        long long x = a[i], y = a[j], cnt = 2;
        v.clear();
        v.push_back(x);
        v.push_back(y);
        mp[x]--;
        mp[y]--;
        while (x + y <= maxx && mp[x + y] > 0) {
          long long tmp = x + y;
          mp[x + y]--;
          v.push_back(x + y);
          x = y;
          y = tmp;
          cnt++;
        }
        for (int i = 0; i < v.size(); i++) mp[v[i]]++;
        ans = max(ans, cnt);
      }
    }
    cout << max(ans, res) << endl;
  }
}
