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[1001];
set<int> S;
map<int, int> countOf;
int n, x;
int temp[1001], tmp = 0;
int check(int i, int j) {
  int f = arr[i], s = arr[j];
  if (f == 0 && s == 0) return countOf[0];
  tmp = 0;
  temp[tmp++] = f;
  temp[tmp++] = s;
  countOf[f]--;
  countOf[s]--;
  int terms = 2;
  while (1) {
    int newter = f + s;
    if (countOf[newter] == 0) break;
    countOf[newter]--;
    temp[tmp++] = newter;
    f = s;
    s = newter;
    terms++;
  }
  for (int i = 0; i < tmp; i++) countOf[temp[i]]++;
  return terms;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  ;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> arr[i];
    countOf[arr[i]]++;
    S.insert(arr[i]);
  }
  sort(arr, arr + n);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      ans = max(check(i, j), ans);
    }
  }
  cout << ans << endl;
  return 0;
}
