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 main() {
  int n;
  map<long long, long long> mp;
  scanf("%d", &n);
  long long x;
  long long t[1001];
  for (int i = 0; i < n; i++) {
    scanf("%I64d", &t[i]);
    mp[t[i]]++;
  }
  int maxima = 0;
  int counter = 0;
  set<pair<long long, long long>> st;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      st.insert(make_pair(t[i], t[j]));
    }
  }
  for (auto iter = st.begin(); iter != st.end(); iter++) {
    counter = 2;
    long long f1 = (*iter).first;
    long long f2 = (*iter).second;
    vector<long long> tempo;
    mp[f1]--;
    mp[f2]--;
    tempo.push_back(f1);
    tempo.push_back(f2);
    for (int k = 0; k < n - 2; k++) {
      if (mp[f1 + f2] > 0) {
        tempo.push_back(f1 + f2);
        long long f3 = f1 + f2;
        f1 = f2;
        f2 = f3;
        mp[f3]--;
        counter++;
      } else
        break;
    }
    for (int i = 0; i < tempo.size(); i++) {
      mp[tempo[i]]++;
    }
    maxima = max(maxima, counter);
  }
  printf("%d", maxima);
  return 0;
}
