Almost Arithmetical Progression

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 indexesi1, 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 in
4000
tegers b1, b2, ..., bn(1 ≤ bi ≤ 106).

Output

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

Sample Input

Input

2
3 5

Output

2

Input

4
10 20 10 30

Output

3

Hint

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

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

/*求最长的子序列，满足隔位的两个数相等*/

#include <cstdio>

#include <algorithm>

using namespace std;

const int MAXN = 4000 + 10;

int n;

int b[MAXN];

int dp[MAXN][MAXN];//定义状态dp[i][j]代表以第i个数为末尾，第j个数为倒数第二个的情况下的最长子序列。

int main()

{

    while (scanf("%d", &n) == 1)

    {

      for (int i = 1; i <= n; ++i)

    {

        scanf("%d", &b[i]);

    }

    int ret = 0;

    dp[0][0] = 0;

    for (int j = 1; j <= n; ++j)

    {

        for (int i = 0, last = 0; i < j; ++i)

        {

            dp[i][j] = dp[last][i] + 1;

            if (b[i] == b[j])

            {

                last = i;

            }

            ret = max(ret, dp[i][j]);

        }

    }

    printf("%d\n", ret);

    }

    return 0;

}