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;
const int N = 1e3 + 5;
const int MOD = 1e9 + 7;
map<int, int> M, m;
const long long int MAX = 1e9;
long long int A[N];
int n;
void rebuild() {
  M.clear();
  for (int i = 1; i <= n; i++) {
    M[A[i]]++;
  }
}
long long int solve(long long int x, long long int y) {
  long long int cnt = 2;
  long long int c = y;
  long long int ox = x, oy = y;
  M[x]--;
  M[y]--;
  while (abs(c) <= MAX) {
    c = x + y;
    x = y;
    y = c;
    if (M[c]) {
      M[c]--;
      cnt++;
    } else
      break;
  }
  x = ox;
  y = oy;
  M[x]++;
  M[y]++;
  c = x + y;
  int curr = 2;
  while (abs(c) <= MAX) {
    c = x + y;
    x = y;
    y = c;
    if (curr < cnt) {
      curr++;
      M[c]++;
    } else
      break;
  }
  return cnt;
}
int main(int argc, char const *argv[]) {
  cin.sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> A[i];
    M[A[i]]++;
    m[A[i]]++;
  }
  long long int ans = 2;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= n; j++) {
      if (i == j) continue;
      long long int f = A[i], s = A[j], next = s;
      if ((!f) && (!s)) {
        ans = max(ans, (long long int)M[0]);
        continue;
      }
      long long int now = solve(f, s);
      ans = max(ans, now);
    }
  }
  cout << ans << '\n';
  return 0;
}
