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 lenLongestFibSubseq(int A[], int n) {
  unordered_map<int, int> mp;
  for (int i = 0; i < n; i++) mp[A[i]]++;
  int ans = 0, x, y;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      x = A[i];
      y = A[j];
      int temp = 0;
      if (x == 0 && y == 0) {
        temp = mp[x];
        ans = max(ans, temp);
        break;
      }
      mp[x]--;
      mp[y]--;
      while (mp[x + y] > 0) {
        mp[x + y]--;
        temp++;
        int z = x;
        x = y;
        y = z + y;
      }
      ans = max(ans, temp + 2);
      mp[x]++;
      mp[y]++;
      while (temp--) {
        int z = y - x;
        mp[z]++;
        y = x;
        x = z;
      }
    }
  }
  return ans;
}
int main() {
  int n;
  cin >> n;
  int A[n + 1];
  for (int i = 0; i < n; i++) cin >> A[i];
  cout << lenLongestFibSubseq(A, n);
  return 0;
}
