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 Set(int N, int pos) { return N = N | (1 << pos); }
int reset(int N, int pos) { return N = N & ~(1 << pos); }
bool check(int N, int pos) { return (bool)(N & (1 << pos)); }
inline int addmod(int x, int y) {
  return (x % 1000000007 + y % 1000000007) % 1000000007;
}
inline int submod(int x, int y) {
  return (x % 1000000007 - y % 1000000007 + 1000000007) % 1000000007;
}
inline int mulmod(int x, int y) {
  return (x % 1000000007 * 1LL * y % 1000000007) % 1000000007;
}
inline int nextSubMask(int i, int mask) { return (i - 1) & mask; }
void we_r_done() {
  cout << "NO";
  exit(0);
}
void debug(bool ok, string mssg) {
  if (ok) return;
  cout << mssg;
  exit(0);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long int f[600];
  f[1] = f[2] = 1;
  for (int i = 3; i <= 50; i++) {
    f[i] = f[i - 1] + f[i - 2];
  }
  int n;
  scanf("%d", &n);
  long long int a[n];
  map<long long int, int> cnt;
  for (int i = 0; i < n; i++) {
    scanf("%lld", &a[i]);
    cnt[a[i]]++;
  }
  int ans = 2;
  map<long long int, bool> done;
  for (int i = 0; i < n; i++) {
    if (done[a[i]]) continue;
    done[a[i]] = true;
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      f[1] = a[i];
      f[2] = a[j];
      if (a[i] == 0 && a[j] == 0) {
        ans = cnt[0];
        continue;
      }
      map<long long int, int> now;
      now[a[i]]++;
      now[a[j]]++;
      for (int len = 3;; len++) {
        f[len] = f[len - 1] + f[len - 2];
        if (now[f[len]] < cnt[f[len]]) {
          ans = max(ans, len);
          now[f[len]]++;
        } else
          break;
      }
    }
  }
  printf("%d", ans);
  return 0;
}
