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 main() {
  int n;
  cin >> n;
  long long int arr[n];
  unordered_map<long long int, long long int> u;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    u[arr[i]]++;
  }
  long long int ans = 0, cnt = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i != j) {
        long long int a = arr[i], b = arr[j];
        if (a == 0 && b == 0) {
          ans = max(ans, u[0]);
          break;
        } else {
          cnt = 2;
          u[arr[i]]--;
          u[arr[j]]--;
          while (1) {
            long long int x = u[a + b];
            if (x > 0) {
              u[a + b]--;
              cnt++;
              long long int temp = b;
              b += a;
              a = temp;
            } else
              break;
          }
          u[b]++;
          u[a]++;
          long long int y = cnt - 2;
          while (y > 0) {
            u[b - a]++;
            long long int temp = a;
            a = b - a;
            b = temp;
            y--;
          }
        }
      }
      ans = max(ans, cnt);
    }
  }
  cout << ans;
  return 0;
}
