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 = 1000000 + 5;
int a[1005], b[1005], sum[1005], pos[1005];
int main() {
  int res = 0, mx = -1, n, l = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    mx = max(a[i], mx);
    if (a[i] == 0) ++res;
  }
  sort(a + 1, a + 1 + n);
  b[++l] = a[1];
  for (int i = 2; i <= n; ++i)
    if (a[i] != a[i - 1]) b[++l] = a[i];
  for (int i = 1; i <= n; ++i) {
    pos[i] = lower_bound(b + 1, b + 1 + l, a[i]) - b;
    ++sum[pos[i]];
  }
  int ans = 2;
  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      if (j == i) continue;
      if (a[i] == 0 && a[j] == 0) continue;
      int x = a[i], y = a[j], cnt = 2;
      vector<int> t;
      t.clear();
      sum[pos[i]]--;
      sum[pos[j]]--;
      t.push_back(pos[i]);
      t.push_back(pos[j]);
      while (x + y <= mx) {
        int z = lower_bound(b + 1, b + 1 + l, x + y) - b;
        if (z == l + 1 || b[z] != x + y || !sum[z]) break;
        sum[z]--;
        t.push_back(z);
        int tmp = x + y;
        x = y;
        y = tmp;
        ++cnt;
      }
      for (int k = 0; k < t.size(); k++) sum[t[k]]++;
      ans = max(ans, cnt);
    }
  }
  printf("%d\n", max(ans, res));
  return 0;
}
