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, vet[1010], val[1010], aux[1010], ptr[1010], qt[1010], cnt0, ans, tt,
    apaga[1010];
int get(int v) { return lower_bound(aux, aux + n, v) - aux; }
int main() {
  scanf("%d", &n);
  for (int(i) = (0); (i) < (n); (i)++)
    scanf("%d", vet + i), aux[i] = vet[i], cnt0 += !vet[i];
  ans = cnt0;
  sort(aux, aux + n);
  for (int(i) = (0); (i) < (n); (i)++) vet[i] = get(vet[i]), qt[vet[i]]++;
  for (int(i) = (0); (i) < (n); (i)++)
    for (int(j) = (0); (j) < (n); (j)++)
      if (i != j && (aux[vet[i]] || aux[vet[j]])) {
        int cur = 2;
        int a = vet[i], b = vet[j];
        apaga[tt++] = a;
        apaga[tt++] = b;
        ptr[a]++;
        ptr[b]++;
        while (1) {
          int c = get(aux[a] + aux[b]);
          if (c == n || aux[c] != aux[a] + aux[b] || ptr[c] == qt[c]) break;
          ptr[c]++;
          a = b;
          b = c;
          cur++;
          apaga[tt++] = c;
        }
        for (int(k) = (0); (k) < (tt); (k)++) ptr[apaga[k]] = 0;
        tt = 0;
        ans = max(ans, cur);
      }
  printf("%d\n", ans);
}
