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 long long MAXN = 1e6 + 10;
long long N, M, ans, val[MAXN];
unordered_map<long long, long long> m;
long long solve(long long, long long);
int main() {
  scanf("%lld", &N);
  for (long long i = 1; i <= N; i++) scanf("%lld", val + i);
  for (long long i = 1; i <= N; i++) m[val[i]]++;
  sort(val + 1, val + N + 1);
  ans = max(ans, m[0]);
  for (long long i = 1; i <= N; i++) {
    for (long long j = 1; j <= N; j++) {
      if (i == j) continue;
      if (val[i] == 0 && val[j] == 0) continue;
      if (val[i] == val[i - 1]) continue;
      m[val[i]]--, m[val[j]]--;
      ans = max(ans, solve(val[i], val[j]) + 2);
      m[val[i]]++, m[val[j]]++;
    }
  }
  printf("%lld\n", ans);
  return 0;
}
long long solve(long long a, long long b) {
  long long ret = 0;
  if (m[a + b]) {
    m[a + b]--;
    ret = solve(b, a + b) + 1;
    m[a + b]++;
  }
  return ret;
}
