C. Almost Arithmetical Progression（DP）

C. Almost Arithmetical Progression

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Gena loves sequences of numbers. Recently, he has discovered a new type of sequences which he called an almost arithmetical progression. A sequence is an almost arithmetical progression, if its elements can be represented as:

• a1 = p, where p is some integer;
• ai = ai - 1 + ( - 1)i + 1·q (i > 1), where q is some integer.

Right now Gena has a piece of paper with sequence b, consisting of n integers. Help Gena, find there the longest subsequence of integers that is an almost arithmetical progression.

Sequence s1,  s2,  ...,  sk is a subsequence of sequence b1,  b2,  ...,  bn, if there is such increasing sequence of indexes i1, i2, ..., ik(1  ≤  i1  <  i2  < ...   <  ik  ≤  n), that bij  =  sj. In other words, sequence s can be obtained from b by crossing out some elements.

Input

The first line contains integer n (1 ≤ n ≤ 4000). The next line contains n integers b1, b2, ..., bn (1 ≤ bi ≤ 106).

Output

Print a single integer — the length of the required longest subsequence.

Examples

input

Copy

2
3 5

output

Copy

2

input

Copy

4
10 20 10 30

output

Copy

3

Note

In the first test the sequence actually is the suitable subsequence.

In the second test the following subsequence fits: 10, 20, 10.

【思路】

dp[i][j] i表示当前数的位置，j表示前面的数。

AC代码：

#include <iostream>
#include <cstring>
using namespace std;
int b[4005];
int dp[4005][4005];
int N;
int main()
{
    cin>>N;
    for(int i=1;i<=N;i++)
    {
        cin>>b[i];
    }
    memset(dp,0,sizeof(dp));
    int res=0;
    for(int i=1;i<=N;i++)
    {
        int t=0;
        for(int j=0;j<i;j++)
        {
            dp[i][j]=dp[j][t]+1;
            if(b[i]==b[j])
                t=j;
            res=max(res,dp[i][j]);
        }
    }
    cout<<res<<endl;
    return 0;
}