题干：

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

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.

题目大意：

   给出一个序列，求最长的子序列，满足隔位的两个数相等，问这个最长的子序列的长度是多少。

解题报告：

  想了好久想到一个dp方程，抱着冒险的心态o(n^2)交一发，TLE41了，，，把离散化的数组提前预处理了一下，，，又莽了一发，，AC了。。小惊喜。

AC代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define ll long long
#define pb push_back
#define pm make_pair
#define fi first
#define se second
using namespace std;
const int MAX = 1e6+5;
int dp[4004][4004];
int a[4004],b[4004];
int cnt[MAX];
int pos[MAX];
int len,n;
int get(int x) {return lower_bound(b+1,b+len+1,x) - b;
}
int main()
{cin>>n;for(int i = 1; i<=n; i++) scanf("%d",a+i),b[i] = a[i],cnt[a[i]]++;sort(b+1,b+n+1);len = unique(b+1,b+n+1) - b - 1;for(int i = 1; i<=n; i++) pos[a[i]] = get(a[i]);for(int i = 1; i<=n; i++) for(int j = 1; j<=n; j++)dp[pos[a[i]]][pos[a[j]]] = 1;for(int i = 1; i<=n; i++) {for(int j = 1; j<i; j++) {if(a[i]!=a[j])dp[pos[a[i]]][pos[a[j]]] = dp[pos[a[j]]][pos[a[i]]]+1;}}int maxx = -1;for(int i = 1; i<=n; i++) {for(int j = 1; j<=n; j++) {maxx = max(maxx,dp[pos[a[i]]][pos[a[j]]]);}}for(int i = 1; i<=n; i++) {maxx = max(maxx,cnt[a[i]]);}printf("%d\n",maxx);return 0 ;}

贴一个题解

再贴一个题解

总结：

   思维过程是这样的，，刚开始想暴力（用vector存权值啊之类的乱搞），，一想肯定不行，，有太多重复操作了。。然后想各种优化后来发现就是要求个序列，，所以跟最长上升子序列联系起来了，，后来发现还真可以写出来个递推式，，但是是o(n^2)的不知道会不会T，，而且1600W的数组大小不知道会不会MLE，，交一发，WA，，造样例，，发现全是同一个数的时候就会爆炸，，我就想，，再多也不可能超过数组长度啊，，所以就像把元素相等的时候特判一下（并没想过正确性），改代码，，测样例，，没啥问题，，再交一发，TLE41，，，失望（但是还是有点小惊喜的因为跑起来了最起码说明算法的问题不大）。再想优化，开个数组预处理一波，去掉了时间复杂度的log，再交，AC。