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 inf = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int maxn = 1e3 + 10;
const int maxv = 1e3 + 10;
const double eps = 1e-9;
int len;
int a[maxn];
int res[maxn];
map<int, int> mp;
map<pair<int, int>, int> vis;
int cnt;
void dfs(int l, int r) {
  vis[pair<int, int>(l, r)] = 1;
  if (mp[l + r] > 0) {
    mp[l + r]--;
    cnt++;
    dfs(r, l + r);
    mp[l + r]++;
  }
}
int main() {
  int n;
  while (scanf("%d", &n) != EOF) {
    mp.clear();
    vis.clear();
    for (int i = 1; i <= n; i++) {
      int tmp;
      scanf("%d", &a[i]);
      mp[a[i]]++;
    }
    sort(a + 1, a + 1 + n);
    int ans = 2;
    for (int i = 1; i <= n; i++) {
      for (int j = 1; j <= n; j++) {
        if (i == j) continue;
        if (vis.count(pair<int, int>(a[i], a[j]))) continue;
        cnt = 2;
        mp[a[i]]--;
        mp[a[j]]--;
        dfs(a[i], a[j]);
        mp[a[i]]++;
        mp[a[j]]++;
        ans = max(ans, cnt);
      }
    }
    cout << ans << endl;
  }
  return 0;
}
