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 int N = 1000000 + 5;
map<long long, int> mp;
long long a[1005];
int main() {
  int n;
  long long res = 0, mx = -1;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%I64d", &a[i]);
    mx = max(a[i], mx);
    if (a[i] == 0) ++res;
    mp[a[i]]++;
  }
  long long ans = 2;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (j == i) continue;
      if (a[i] == 0 && a[j] == 0) continue;
      long long x = a[i], y = a[j], cnt = 2;
      vector<long long> t;
      t.clear();
      t.push_back(x);
      t.push_back(y);
      mp[x]--;
      mp[y]--;
      while (x + y <= mx && mp[x + y] > 0) {
        long long tmp = x + y;
        mp[tmp]--;
        t.push_back(tmp);
        x = y;
        y = tmp;
        ++cnt;
      }
      for (int k = 0; k < t.size(); k++) mp[t[k]]++;
      ans = max(ans, cnt);
    }
  }
  printf("%I64d\n", max(ans, res));
  return 0;
}
