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() {
  long long int n;
  cin >> n;
  long long int arr[n];
  unordered_map<long long int, long long int> umap;
  for (long long int i = 0; i < n; i++) {
    cin >> arr[i];
    umap[arr[i]]++;
  }
  long long int ans = 0;
  for (long long int i = 0; i < n; i++) {
    long long int a = arr[i];
    for (long long int j = 0; j < n; j++) {
      long long int b = arr[j];
      if (i == j) continue;
      if (arr[i] == arr[j] && arr[j] == 0) {
        ans = max(ans, umap[arr[i]]);
        continue;
      }
      long long int cnt = 2;
      long long int c = a + b;
      umap[a]--;
      umap[b]--;
      while (umap.find(c) != umap.end() && umap[c] != 0) {
        umap[c]--;
        a = b;
        b = c;
        c = a + b;
        cnt++;
      }
      ans = max(ans, cnt);
      while (cnt > 2) {
        a = b - a;
        b = b - a;
        c = a + b;
        umap[c]++;
        cnt--;
      }
      umap[arr[i]]++;
      umap[arr[j]]++;
    }
  }
  cout << ans << endl;
  return 0;
}
