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;
unordered_map<int, int> S;
int n;
int a[1010];
int f[2];
int ans;
vector<int> N;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    S[a[i]]++;
  }
  for (auto i : S) {
    N.push_back(i.first);
    if (i.second >= 2) N.push_back(i.first);
  }
  n = N.size();
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      if (i != j) {
        unordered_map<int, int> T;
        f[0] = N[i];
        f[1] = N[j];
        T[f[0]]++;
        int t = 1, tans = 1;
        while (S[f[t]] - T[f[t]]) {
          T[f[t]]++;
          tans++;
          t ^= 1;
          f[t] = f[t] + f[t ^ 1];
        }
        ans = max(ans, tans);
      }
  printf("%d", ans);
}
