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;
int n;
long long a[1010];
long long ans;
long long vis[1010];
map<long long, long long> mp;
void dfs(long long x, long long y, long long cnt) {
  ans = max(cnt, ans);
  if (mp[x + y]) {
    mp[x + y]--;
    dfs(y, x + y, cnt + 1);
    mp[x + y]++;
  }
}
int main() {
  while (cin >> n) {
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i <= n; i++) mp[a[i]]++;
    long long kk = 0;
    for (int i = 1; i <= n; i++)
      if (a[i] == 0) kk++;
    ans = 0;
    for (int i = 0; i <= n; i++) vis[i] = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        if (i != j) {
          if (a[i] == 0 && a[j] == 0) {
            ans = max(ans, kk);
            continue;
          }
          mp[a[i]]--;
          mp[a[j]]--;
          dfs(a[i], a[j], 2);
          mp[a[i]]++, mp[a[j]]++;
        }
      }
    }
    cout << ans << endl;
  }
  return 0;
}
