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;
map<int, int> m;
map<pair<int, int>, int> m2;
int n, a[1010], ans = -1;
void dfs(int x, int y, int l) {
  if (m[x + y] == 0) {
    if (l > ans) ans = l;
    return;
  }
  m[x + y]--;
  dfs(y, x + y, l + 1);
  m[x + y]++;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    m[a[i]]++;
  }
  sort(a + 1, a + n + 1);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      if (m2[make_pair(a[i], a[j])] == 1) continue;
      m2[make_pair(a[i], a[j])] = 1;
      m[a[i]]--;
      m[a[j]]--;
      dfs(a[i], a[j], 2);
      m[a[i]]++;
      m[a[j]]++;
    }
  }
  printf("%d", ans);
  return 0;
}
