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;
template <typename T>
void read(T &x) {
  x = 0;
  int f = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -1;
  for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
  x = x * f;
}
int n, sav, d[1020], f[1020], cnt, v, ans = -1;
map<int, int> mp;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> sav;
    d[i] = sav;
    mp[sav]++;
  }
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      if (d[i] == 0 and d[j] == 0) {
        ans = max(ans, mp[0]);
        continue;
      }
      cnt = 2;
      f[1] = d[i];
      f[2] = d[j];
      v = f[1] + f[2];
      mp[d[i]]--;
      mp[d[j]]--;
      while (mp[v] > 0) {
        mp[v]--;
        cnt++;
        f[cnt] = v;
        v = f[cnt - 1] + f[cnt];
      }
      ans = max(cnt, ans);
      for (int k = 1; k <= cnt; k++) mp[f[k]]++;
    }
  cout << ans << endl;
  return 0;
}
