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>
int read() {
  register char ch = getchar();
  register int ret = 0;
  while (!isdigit(ch)) ch = getchar();
  while (isdigit(ch)) {
    ret = ret * 10 + ch - '0';
    ch = getchar();
  }
  return ret;
}
int n, k, d, ans;
long long a[500055], b[500055];
bool f[500055];
std::map<long long, int> s;
int main() {
  n = read();
  for (register int i = (1); i <= (n); ++i) {
    scanf("%lld", &a[i]);
    s[a[i]]++;
  }
  std::sort(a + 1, a + n + 1);
  ans = 2;
  for (register int i = (1); i <= (n); ++i)
    for (register int j = (1); j <= (n); ++j)
      if (i != j) {
        b[1] = a[i];
        b[2] = a[j];
        int p = 3;
        if (b[1] == 0 && b[2] == 0) {
          ans = std::max(ans, s[0]);
          continue;
        }
        s[b[1]]--;
        s[b[2]]--;
        while (p <= n && s[b[p - 1] + b[p - 2]]) {
          b[p] = b[p - 1] + b[p - 2];
          s[b[p]]--;
          ++p;
        }
        --p;
        ans = std::max(ans, p);
        while (p) s[b[p--]]++;
      }
  printf("%d", ans);
  return 0;
}
