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 a[1000], n;
map<int, int> mp;
int BinSearch(int val) {
  int l = 0, r = n;
  while (l < r) {
    int mid = (l + r) / 2;
    if (a[mid] >= val)
      r = mid;
    else
      l = mid + 1;
  }
  return l;
}
int solve(int prv, int cur) {
  long long x = (long long)a[prv] + a[cur];
  int ans = 0;
  if (mp[x] > 0) {
    mp[x]--;
    ans = solve(cur, BinSearch(x)) + 1;
    mp[x]++;
  } else
    return 0;
  return ans;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
    mp[a[i]]++;
  }
  sort(a, a + n);
  int ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i != j) {
        if (a[i] == a[j] && a[i] == 0) continue;
        mp[a[i]]--;
        mp[a[j]]--;
        ans = max(ans, solve(i, j));
        mp[a[i]]++;
        mp[a[j]]++;
      }
    }
  }
  cout << max(ans, mp[0] - 2) + 2 << endl;
  return 0;
}
