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 N = 1005;
int a[N];
multiset<int> values;
int main() {
  int t = 1, n;
  while (t--) {
    scanf("%d", &(n));
    for (int i = 1; i <= n; i++) {
      scanf("%d", &(a[i]));
      values.insert(a[i]);
    }
    int c0 = values.count(0), ans = 2;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        if (i == j) continue;
        int x = a[i], y = a[j];
        if (x == 0 && y == 0) {
          ans = max(ans, c0);
          continue;
        }
        int l = 2;
        vector<int> deletions;
        deletions.push_back(x);
        deletions.push_back(y);
        values.erase(values.find(x));
        values.erase(values.find(y));
        while (1) {
          int temp = y;
          y += x;
          x = temp;
          if (y > 1e9 || values.find(y) == values.end()) break;
          l++;
          if (l <= 10) {
            values.erase(values.find(y));
            deletions.push_back(y);
          }
        }
        for (int x : deletions) values.insert(x);
        ans = max(ans, l);
      }
    }
    printf("%d\n", ans);
  }
}
