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 n, m, z, w, tmp, ans;
long long A[1005], x, y, k;
map<long long, int> M, P;
map<int, int> Q;
int main() {
  scanf("%d", &n);
  for (int i = (0); i < (n); i++) {
    scanf("%I64d", &A[i]);
    M[A[i]]++;
    P[A[i]]++;
  }
  sort(A, A + n);
  for (int i = (0); i < (n); i++) {
    if (i != 0) {
      if (A[i] == A[i - 1]) continue;
    }
    for (int j = (0); j < (n); j++)
      if (i != j) {
        if (j != 0) {
          if (A[j] == A[j - 1] && j - 1 != i) continue;
        }
        if (A[i] == 0ll && A[j] == 0ll) continue;
        x = A[i];
        y = A[j];
        M[x]--;
        M[y]--;
        tmp = 2;
        while (M[x + y] > 0) {
          M[x + y]--;
          k = x + y;
          x = y;
          y = k;
          tmp++;
        }
        ans = max(ans, tmp);
        x = A[i];
        y = A[j];
        M[x]++;
        M[y]++;
        while (P[x + y] != M[x + y]) {
          M[x + y]++;
          k = x + y;
          x = y;
          y = k;
        }
      }
  }
  tmp = 0;
  for (int i = (0); i < (n); i++)
    if (A[i] == 0ll) tmp++;
  ans = max(ans, tmp);
  printf("%d\n", ans);
  return 0;
}
