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 n, ans = 2, cnt;
int a[1111];
map<int, int> all;
vector<int> used;
inline void solve(int x, int y) {
  long long sum = x + y;
  used.push_back(x), used.push_back(y);
  all[x]--;
  all[y]--;
  cnt = 2;
  while (all[sum]) {
    cnt++;
    all[sum]--;
    used.push_back(sum);
    long long x = sum;
    sum = sum + y;
    y = x;
    ans = max(ans, cnt);
  }
  for (int i = 0; i < (int)used.size(); i++) {
    all[used[i]]++;
  }
  used.clear();
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    all[a[i]]++;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (a[i] == 0 && a[j] == 0)
        ans = max(ans, all[0]);
      else {
        solve(a[i], a[j]);
        solve(a[j], a[i]);
      }
    }
  }
  cout << ans;
  return 0;
}
