Description：
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 nintegers b1, b2, ..., bn (1 ≤ bi ≤ 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.
把上面的两个式子化简以后就会得到 ，p，p-q,p,p-q,p,p-q.......这样的一个序列，然后让球符合的最大长度，就是求每隔一个数相等的最大长度 ，就类似1,2,1,2,1,2,1,2这样一个序列，
用一个二维数组表示dp【i】【j】表示最后一位是a【i】倒数第二位是a【j】的最大长度
AC代码：
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#include<stdlib.h>
#include<queue>
#include<map>
#include<set>
#include<iomanip>
#include<math.h>
using namespace std;
typedef long long ll;
typedef double ld;
const int INF = 0x3f3f3f3f;
const int N=5000;
int i,j,k;
int m,n,t;
int p,q;
int l,r;
int res,cnt,ans,temp,x;
int a[N];
int dp[N][N];
int main()
{
scanf("%d",&n);
for(i=1; i<=n; i++)
scanf("%d",&a[i]);
memset(dp,0,sizeof(dp));
ans=-1;
for(i=1; i<=n; i++)
{
cnt=0;
for(j=0; j<i; j++)
{
dp[i][j]=dp[j][cnt]+1;
if(a[i]==a[j])
cnt=j;
ans=max(ans,dp[i][j]);
}
}
printf("%d\n",ans);
return 0;
}
