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 + 10, LG = 20, M = 1e4 + 10;
int a[N], MAXI = -1e9 - 10, MINI = 1e9 + 10;
unordered_map<int, int> mp;
int main() {
  int n, ans = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
    mp[a[i]]++;
    MAXI = max(MAXI, a[i]);
    MINI = min(MINI, a[i]);
  }
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      if (i == j) continue;
      if (a[i] == a[j] && a[i] == 0) {
        ans = max(ans, mp[0]);
        continue;
      }
      vector<int> t;
      mp[a[i]]--;
      mp[a[j]]--;
      int val = a[i] + a[j], cnt = 0;
      int x = a[i], y = a[j];
      t.push_back(x);
      t.push_back(y);
      while (val >= MINI && val <= MAXI) {
        if (mp[val] > 0)
          cnt++;
        else
          break;
        mp[val]--;
        x = y;
        y = val;
        t.push_back(val);
        val = x + y;
      }
      ans = max(ans, cnt + 2);
      for (int k = 0; k < cnt + 2; k++) mp[t[k]]++;
    }
  }
  printf("%d\n", ans);
}
