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;
void fre() {
  freopen("c://test//input.in", "r", stdin);
  freopen("c://test//output.out", "w", stdout);
}
template <class T1, class T2>
inline void gmax(T1 &a, T2 b) {
  if (b > a) a = b;
}
template <class T1, class T2>
inline void gmin(T1 &a, T2 b) {
  if (b < a) a = b;
}
const int N = 1010, M = 0, Z = 1e9 + 7, ms63 = 0x3f3f3f3f;
int n;
int a[N];
int ans;
map<int, int> mop;
long long f[N];
void go(int x, int y) {
  --mop[x];
  --mop[y];
  f[1] = x;
  f[2] = y;
  int i;
  for (i = 3;; ++i) {
    f[i] = f[i - 1] + f[i - 2];
    if (mop.find(f[i]) != mop.end() && mop[f[i]])
      --mop[f[i]];
    else
      break;
  }
  gmax(ans, i - 1);
  for (int j = 1; j < i; ++j) ++mop[f[j]];
}
pair<int, int> b[N * N];
int main() {
  while (~scanf("%d", &n)) {
    mop.clear();
    for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), ++mop[a[i]];
    int m = 0;
    for (int i = 1; i <= n; ++i) {
      for (int j = 1; j <= n; ++j)
        if (j != i) {
          b[++m] = make_pair(a[i], a[j]);
        }
    }
    b[0].first = 2e9;
    sort(b + 1, b + m + 1);
    ans = 0;
    for (int i = 1; i <= m; ++i) {
      if (b[i].first != b[i - 1].first || b[i].second != b[i - 1].second) {
        go(b[i].first, b[i].second);
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}
