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 a[1005];
map<long long, int> mp;
int main() {
  int n;
  long long sum = 0, maxx = 2, mx = -1;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%I64d", &a[i]);
    mx = max(mx, a[i]);
    if (a[i] == 0) sum++;
    mp[a[i]]++;
  }
  maxx = max(maxx, sum);
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      if (a[i] == 0 && a[j] == 0) continue;
      sum = 2;
      long long x = a[i], y = a[j];
      vector<long long> t;
      t.clear();
      t.push_back(x);
      t.push_back(y);
      mp[x]--;
      mp[y]--;
      while (x + y <= mx && mp[x + y] > 0) {
        long long tmp = x + y;
        mp[tmp]--;
        t.push_back(tmp);
        x = y;
        y = tmp;
        sum++;
      }
      maxx = max(maxx, sum);
      for (int k = 0; k < t.size(); k++) mp[t[k]]++;
    }
  }
  printf("%I64d\n", maxx);
  return 0;
}
