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 main() {
  int n;
  scanf("%d", &n);
  int a[n];
  set<int> has;
  int zeros = 0;
  map<int, int> f;
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    zeros += (a[i] == 0);
    f[a[i]]++;
    has.insert(a[i]);
  }
  int ans = max(2, zeros);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i != j) {
        int x = a[i], y = a[j];
        if (x == 0 && y == 0) break;
        map<int, int> cnt;
        cnt[x]++, cnt[y]++;
        int cur = 2;
        while (has.find(x + y) != has.end()) {
          int t = x;
          x = y;
          y = y + t;
          if (++cnt[y] > f[y]) break;
          cur++;
        }
        ans = max(ans, cur);
      }
    }
  }
  printf("%d\n", ans);
}
