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;
long long n, ans, cnt0;
long long a[10010], b[10010], cntb;
map<long long, int> mp;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &a[i]);
    mp[a[i]]++;
    if (!a[i]) cnt0++;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (a[i] == 0 && a[j] == 0) {
        ans = max(ans, cnt0);
        continue;
      }
      if (i != j) {
        cntb = 0;
        long long x = a[i], y = a[j], z;
        mp[x]--;
        mp[y]--;
        b[++cntb] = x;
        b[++cntb] = y;
        long long k = 2;
        while (mp[x + y] > 0) {
          mp[x + y]--;
          long long t = x + y;
          x = y;
          y = t;
          b[++cntb] = t;
          k++;
        }
        ans = max(ans, k);
        for (int p = 1; p <= cntb; p++) {
          mp[b[p]]++;
        }
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}
