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 arr[100001];
map<int, int> aa;
vector<int> c[100];
int lol(int last, int last_last) {
  int ans = 0;
  if (aa[last + last_last]) {
    aa[last + last_last]--;
    ans = max(ans, lol(last_last, last_last + last) + 1);
    aa[last + last_last]++;
  }
  return ans;
}
int main() {
  int a, b, c;
  cin >> a;
  for (int x = 0; x < a; x++) {
    cin >> arr[x];
    aa[arr[x]]++;
  }
  int ans = 0;
  for (int x = 0; x < a; x++) {
    for (int p = 0; p < a; p++) {
      if (p != x && !(arr[x] == arr[p] && arr[x] == 0)) {
        aa[arr[x]]--;
        aa[arr[p]]--;
        ans = max(ans, lol(arr[x], arr[p]) + 2);
        aa[arr[x]]++;
        aa[arr[p]]++;
      }
    }
  }
  cout << max(ans, aa[0]);
}
