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;
const int INF = 0x3f3f3f3f;
map<int, int> num;
int f(int i, int j) {
  int ans = 0;
  if (num[i + j]) {
    --num[i + j];
    ans = f(j, i + j) + 1;
    ++num[i + j];
  }
  return ans;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  cin >> n;
  int a[1000];
  for (int i = 0; i < n; ++i) {
    cin >> a[i];
    ++num[a[i]];
  }
  sort(a, a + n);
  int N = unique(a, a + n) - a;
  int ans = max(2, num[0]);
  for (int i = 0; i < N; ++i) {
    for (int j = 0; j < N; ++j) {
      if (i != j || num[a[i]] > 1) {
        --num[a[i]];
        --num[a[j]];
        ans = max(ans, f(a[i], a[j]) + 2);
        ++num[a[i]];
        ++num[a[j]];
      }
    }
  }
  cout << ans << endl;
  return 0;
}
