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 MAX = 1010;
int arr[MAX], n;
unordered_map<int, int> mp;
set<pair<int, int> > ss;
int solve(int a, int b) {
  auto &it = mp[b];
  it--;
  if (!mp[a + b]) return it++, 2;
  int ret = 1 + solve(b, a + b);
  it++;
  return ret;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", arr + i);
    mp[arr[i]]++;
  }
  int ans = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i != j && ss.insert({arr[i], arr[j]}).second) {
        auto &it = mp[arr[i]];
        it--;
        ans = max(ans, solve(arr[i], arr[j]));
        it++;
      }
  printf("%d\n", ans);
  return 0;
}
