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;
int n, ans = 0;
vector<int> arr;
unordered_map<int, int> mp;
map<pair<int, int>, bool> yes;
void check(int a, int b) {
  a = arr[a];
  b = arr[b];
  if (yes[make_pair(a, b)]) {
    return;
  }
  yes[make_pair(a, b)] = true;
  if (a == 0 && b == 0) {
    return;
  }
  unordered_map<int, int> mm;
  mm[a]++;
  mm[b]++;
  int cur = 2;
  for (int c = a + b; c <= int(1e9) && c >= -int(1e9);) {
    if (mm[c] + 1 > mp[c]) {
      break;
    }
    ++mm[c];
    a = b;
    b = c;
    c = a + b;
    ++cur;
  }
  ans = max(ans, cur);
}
int main(int argc, char *argv[]) {
  scanf("%d", &n);
  arr.resize(n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &arr[i]);
    mp[arr[i]]++;
    if (arr[i] == 0) {
      ++ans;
    }
  }
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      if (i != j) {
        check(i, j);
      }
    }
  }
  printf("%d", ans);
  return 0;
}
