Manthan, Codefest 16 D. Fibonacci-ish

D. Fibonacci-ish

time limit per test

3 seconds

memory limit per test

512 megabytes

input

standard input

output

standard output

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 , , , , 28.

思路：直接暴力枚举前两个元素，然后先离散化下，并用map一一对应下数值，开个数组记录各个值的个数，然后根据斐波那契数列依次推下个数字

在数组中查找，其中开始为0，0需要特判否则复杂度n³；

 1 #include<stdio.h>
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<string.h>
 5 #include<math.h>
 6 #include<queue>
 7 #include<map>
 8 using namespace std;
 9 const long long N=1e9+7;
10 typedef long long LL;
11 LL vv[25];
12 LL MM[1005];
13 int NN[1005];
14 int KK[1005];
15 map<LL,LL>my;
16 LL quickmi(long long  a,long long  b);
17 int main(void)
18 {
19     int i,j,k;
20     while(scanf("%d",&k)!=EOF)
21     {memset(NN,0,sizeof(NN));
22         for(i=0; i<k; i++)
23         {
24             scanf("%I64d",&MM[i]);
25         }
26         int cnt=1;sort(MM,MM+k);my[MM[0]]=1;
27         NN[1]++;
28         for(i=1; i<k; i++)
29         {
30             if(MM[i]!=MM[i-1])
31                 cnt++;
32             NN[cnt]++;
33             my[MM[i]]=cnt;
34         }int maxx=2;
35         for(i=0; i<k; i++)
36         {
37             for(j=0; j<k; j++)
38             {int sum=2;
39                 if(i!=j)
40                 {int KK[1005]={0};LL p=MM[i];LL q=MM[j];
41                 if(p==0&&q==0)
42                 {
43                     sum=NN[my[0]];
44                 }
45                 else {KK[my[p]]++;KK[my[q]]++;
46                  while(true)
47                  {
48                      LL ans=p+q;
49                      KK[my[ans]]++;
50                      if(KK[my[ans]]<=NN[my[ans]])
51                      sum++;
52                      else break;
53                      p=q;q=ans;
54                  }}
55                  if(sum>maxx)
56                     maxx=sum;}
57             }
58         }printf("%d\n",maxx);
59     }
60     return 0;
61 }
62
63 LL quickmi(long long  a,long long  b)
64 {
65     LL sum=1;
66     while(b)
67     {
68         if(b&1)
69             sum=(sum*a)%(N);
70         a=(a*a)%N;
71         b/=2;
72     }
73     return sum;
74 }