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;
int a[1100], n, del[100], lend;
map<long long, int> m;
int main() {
  while (~scanf("%d", &n)) {
    int ans = 2;
    m.clear();
    for (int i = 0; i < n; ++i) {
      scanf("%d", a + i);
      m[a[i]]++;
    }
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < n; ++j) {
        if (i == j) continue;
        int x = a[i], y = a[j], sum = a[i] + a[j], len = 2;
        lend = 0;
        if (x == 0 && y == 0) {
          ans = max(ans, m[0]);
          continue;
        }
        del[lend++] = a[i];
        del[lend++] = a[j];
        m[a[i]]--;
        m[a[j]]--;
        while (m[sum] != 0) {
          m[sum]--;
          del[lend++] = sum;
          len++;
          x = y;
          y = sum;
          sum = x + y;
        }
        for (int k = 0; k < lend; ++k) {
          m[del[k]]++;
        }
        ans = max(ans, len);
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}
