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 i, j, m, n, p, k, a[1005], ans, Q[1005], l;
vector<int> v[100007];
vector<int> sum[100007];
void del(int x) {
  int p = abs(x) % 100007, j;
  for (j = 0; j < (int)v[p].size(); ++j)
    if (v[p][j] == x) break;
  sum[p][j]--;
}
void ins(int x) {
  int p = abs(x) % 100007, j;
  for (j = 0; j < (int)v[p].size(); ++j)
    if (v[p][j] == x) break;
  sum[p][j]++;
}
int main() {
  scanf("%d", &n);
  for (i = 1; i <= n; ++i) scanf("%d", &a[i]), ans += (a[i] == 0);
  for (i = 1; i <= n; ++i) {
    int p = abs(a[i]) % 100007, S = v[p].size();
    for (j = 0; j < S; ++j)
      if (v[p][j] == a[i]) break;
    if (j == S)
      v[p].push_back(a[i]), sum[p].push_back(1);
    else
      sum[p][j]++;
  }
  for (i = 1; i <= n; ++i)
    for (j = 1; j <= n; ++j)
      if (i != j) {
        if (a[i] == 0 && a[j] == 0) continue;
        int A = a[i], B = a[j];
        del(A);
        del(B);
        Q[Q[0] = 1] = A, Q[++Q[0]] = B;
        for (l = 2;; ++l) {
          int C = A + B, p = abs(C) % 100007;
          for (k = 0; k < (int)v[p].size(); ++k)
            if (v[p][k] == C) break;
          if (k == v[p].size() || sum[p][k] == 0) break;
          Q[++Q[0]] = C;
          del(C);
          A = B;
          B = C;
        }
        ans = max(ans, l);
        for (k = 1; k <= Q[0]; ++k) ins(Q[k]);
      }
  printf("%d\n", ans);
}
