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[1005];
map<int, int> fab;
int getans(int a, int b) {
  int ans = 0;
  if (fab[a + b]) {
    fab[a + b]--;
    ans = getans(b, a + b) + 1;
    fab[a + b]++;
  }
  return ans;
}
int main() {
  int n, m;
  fab.clear();
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    fab[a[i]]++;
  }
  sort(a, a + n);
  m = unique(a, a + n) - a;
  int ans = 0;
  for (int i = 0; i < m; i++) {
    for (int j = 0; j < m; j++) {
      if (i == j && fab[a[i]] == 1) continue;
      fab[a[i]]--;
      fab[a[j]]--;
      ans = max(ans, getans(a[i], a[j]) + 2);
      fab[a[i]]++;
      fab[a[j]]++;
    }
  }
  printf("%d", ans);
  return 0;
}
