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;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
int n, mx;
int a[1005];
map<int, int> mp;
vector<int> q;
void solve(int x, int y) {
  int ans = 2;
  if (x == 0 && y == 0) return;
  mp[x]--;
  mp[y]--;
  q.push_back(x);
  q.push_back(y);
  while (1) {
    int z = x + y;
    if (mp[z]) {
      ans++;
      mp[z]--;
      q.push_back(z);
    } else
      break;
    x = y;
    y = z;
  }
  mx = max(mx, ans);
  for (int i = 0; i < q.size(); i++) mp[q[i]]++;
  q.clear();
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i <= n; i++) mp[a[i]]++;
  mx = mp[0];
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= n; j++)
      if (i != j) solve(a[i], a[j]);
  printf("%d\n", mx);
  return 0;
}
