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>
#pragma warning(disable : 4996)
#pragma comment(linker, "/STACK:36777216")
using namespace std;
const long double EPS = 1e-9;
const int N = 100000;
const long long int INF = 1E17;
const long long int MOD = 1000000007;
int check(int a) {
  int ans = 0;
  while (a > 0) {
    ans += a / 5;
    a /= 5;
  }
  return ans;
}
int main() {
  clock_t start = clock();
  srand(13);
  cout << fixed << setprecision(10);
  int n;
  cin >> n;
  vector<int> arr(n);
  map<int, int> mp;
  for (int i = 0; i < n; ++i) {
    cin >> arr[i];
    mp[arr[i]]++;
  }
  random_shuffle(arr.begin(), arr.end());
  int ans = 2;
  vector<int> used;
  for (int i = 0; i < n; ++i) {
    if (clock() - start > 2500) break;
    for (int j = 0; j < n; ++j) {
      if (j == i) continue;
      int c = 2;
      int a = arr[i];
      int b = arr[j];
      used.push_back(a);
      mp[a]--;
      if (mp[a] == 0) {
        mp.erase(a);
      }
      used.push_back(b);
      mp[b]--;
      if (mp[b] == 0) {
        mp.erase(b);
      }
      int curr = 2;
      while (mp.count(a + b)) {
        int c = a;
        a = b;
        b = c + b;
        curr++;
        mp[b]--;
        if (mp[b] == 0) {
          mp.erase(b);
        }
        used.push_back(b);
      }
      ans = max(ans, curr);
      for (int i = (used.size()) - 1; i >= 0; --i) {
        mp[used[i]]++;
      }
      used.clear();
    }
  }
  cout << ans;
  return 0;
}
