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>
#pragma warning(disable : 4996)
using namespace std;
int IT_MAX = 131072;
const long long MOD = 1000000009;
const int INF = 1034567891;
const long long LL_INF = 2234567890123456789ll;
const double PI = 3.141592653589793238;
const long double ERR = 1E-10;
int in[100050];
unordered_map<int, int> Mc;
int main() {
  int N, i, j, k, l, cnt0 = 0;
  scanf("%d", &N);
  for (i = 1; i <= N; i++) {
    scanf("%d", &in[i]);
    if (in[i] == 0) cnt0++;
    Mc[in[i]]++;
  }
  int ans = cnt0;
  for (i = 1; i <= N; i++) {
    for (j = 1; j <= N; j++) {
      if (i == j) continue;
      if (in[i] == 0 && in[j] == 0) continue;
      int t1 = in[i], t2 = in[j];
      Mc[t1]--, Mc[t2]--;
      for (k = 2;; k++) {
        if (Mc.count(t1 + t2) == 0 || Mc[t1 + t2] == 0) break;
        Mc[t1 + t2]--;
        t2 = t1 + t2;
        t1 = t2 - t1;
      }
      ans = max(ans, k);
      if (k == 11) {
        i = i;
      }
      t1 = in[i], t2 = in[j];
      Mc[t1]++, Mc[t2]++;
      for (l = 2; l < k; l++) {
        if (Mc.count(t1 + t2) == 0) break;
        Mc[t1 + t2]++;
        t2 = t1 + t2;
        t1 = t2 - t1;
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}
