Almost Arithmetical Progression(dp)

Almost Arithmetical Progression

 CodeForces - 255C 

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;
• a_i = a_i - 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,  ...,  s_k is a subsequence of sequence b₁,  b₂,  ...,  b_n, if there is such increasing sequence of indexes i₁, i₂, ..., i_k (1  ≤  i₁  <  i₂  < ...   <  i_k  ≤  n), that b_ij  =  s_j. 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, ..., b_n (1 ≤ b_i ≤ 10⁶).

Output

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

Examples

Input

2
3 5

Output

2

Input

4
10 20 10 30

Output

3

Note

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

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

题意：找一个最长的锯齿形子串 例如10 30 10 30 10

思路：%大佬%大佬%大佬 大佬tql 

　　  因为数据在1e6而n只有4000，所以大佬说可以先搞一波离散化，然后开二维dp。dp[x][y]中x表示当前数的位置，y表示x这个位置上的数字前面的数，例如1 4 1 4 dp[3][4]代表当数字为1且其前面的数字为4时的最大子串长度。dp[3][4]=dp[2][1]+1,具体看代码。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
const int maxn=4010;
int a[maxn],id[maxn],b[maxn];
int dp[maxn][maxn];
int n;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&a[i]);
        id[i]=a[i];    
    }
    sort(id+1,id+1+n);
    int len=unique(id+1,id+1+n)-(id+1);
    for(int i=1;i<=n;i++)
    {
        a[i]=lower_bound(id+1,id+1+n,a[i])-id;
    }
    int ans=0;
    for(int i=1;i<=n;i++)
    {
        for(int j=1;j<i;j++)
        {
            dp[i][a[j]]=dp[j][a[i]]+1;
            ans=max(ans,dp[i][a[j]]);
        }
    }
    printf("%d\n",ans+1);
    return 0;
}