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 MX = 3 * 1e3 + 100;
const long long INF = 1e18;
int N, M;
long long arr[MX];
map<pair<long long, pair<long long, long long> >, int> memo;
map<long long, int> can;
void del(long long x) {
  if (can[x] > 1)
    can[x]--;
  else
    can.erase(x);
}
int main() {
  cin >> N;
  for (int i = 0; i < N; i++) cin >> arr[i], can[arr[i]]++;
  int best = 2;
  for (int i = 0; i < N; i++) {
    for (int j = 0; j < N; j++) {
      if (j == i) continue;
      long long x = arr[i], y = arr[j], z = x + y;
      del(x);
      del(y);
      if (can.find(z) == can.end()) {
        can[x]++;
        can[y]++;
        continue;
      }
      pair<long long, pair<long long, long long> > cur =
          make_pair(x, make_pair(y, z));
      if (memo.find(cur) != memo.end()) {
        can[x]++;
        can[y]++;
        continue;
      }
      memo[cur]++;
      vector<long long> d;
      d.push_back(x);
      d.push_back(y);
      d.push_back(z);
      del(z);
      int cnt = 3;
      while (cnt < N) {
        x = y;
        y = z;
        z = x + y;
        if (can.find(z) == can.end()) break;
        cnt++;
        d.push_back(z);
        del(z);
      }
      best = max(best, cnt);
      for (int k = 0; k < d.size(); k++) can[d[k]]++;
    }
  }
  cout << best << "\n";
  return 0;
}
