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 7,14,21,35,28.

题意：

斐波拉契数列数列，包括负数，求满足的最大个数。

思路：

总共不会超过90个，不断标记，即可。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <map>
using namespace std;
#define N 1010

map <int, int> mp;
int a[N];
int temp[N];

int main()
{
    int n ;
    cin >> n;
    mp.clear();
    for(int i = 0; i < n; i++){
        scanf("%d", a + i);
        mp[a[i]]++;
        //该数出现的次数
    }
    int ans = 2, now = 0, cnt, x, y, z;
    for(int i = 0; i < n; i++){
        for(int j = 0; j < n; j++){
            if(i == j) continue;
            x = a[i], y = a[j];
            if( x == 0 && y == 0 ){
                //两个都是0的话，直接判断0的个数，因为0+0=0
                if(mp[0] > ans) ans = mp[0];
                continue;
            }
            cnt = 0;
            mp[x]--; mp[y]--;
            temp[cnt++] = x;
            temp[cnt++] = y;
            //先把要加的数从map里面删除，避免重复，用temp存放，之后再放回去
            while(mp[x+y] > 0){
                z = x + y;
                mp[z]--;
                temp[cnt++] = z;
                x = y;
                y = z;
            }
            for(int i = 0; i < cnt; i++)
                mp[temp[i]]++;
            if(cnt > ans) ans = cnt;
        }
    }
    printf("%d\n",ans);
    return 0;
}

代码转自别处，侵权请联系删除。