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 = 1005;
int a[100005];
map<int, int> mp;
map<pair<int, int>, int> mm;
int s;
int dfs(int a, int b) {
  if (a == 0 && b == 0) return mp[0];
  if (mp[a + b] == 0) return 0;
  if (mp[a + b] > 0) {
    mp[a + b]--;
    s = dfs(b, a + b) + 1;
    mp[a + b]++;
    return s;
  }
}
int main() {
  int i, j, n;
  scanf("%d", &n);
  for (i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    mp[a[i]]++;
  }
  int ans = 0;
  for (i = 0; i < n; i++) {
    for (j = 0; j < n; j++) {
      if (i != j) {
        mp[a[j]]--;
        mp[a[i]]--;
        s = 0;
        ans = max(ans, dfs(a[i], a[j]) + 2);
        mp[a[i]]++;
        mp[a[j]]++;
      }
    }
  }
  printf("%d\n", ans);
}
