（dp）CF 813 Educational Codeforces Round 22 D. Two Melodies

D. Two Melodies

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Alice is a beginner composer and now she is ready to create another masterpiece. And not even the single one but two at the same time!

Alice has a sheet with n notes written on it. She wants to take two such non-empty non-intersecting subsequences that both of them form a melody and sum of their lengths is maximal.

Subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements.

Subsequence forms a melody when each two adjacent notes either differs by 1 or are congruent modulo 7.

You should write a program which will calculate maximum sum of lengths of such two non-empty non-intersecting subsequences that both of them form a melody.

Input

The first line contains one integer number n (2 ≤ n ≤ 5000).

The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 105) — notes written on a sheet.

Output

Print maximum sum of lengths of such two non-empty non-intersecting subsequences that both of them form a melody.

Examples

input

4
1 2 4 5

output

4

input

6
62 22 60 61 48 49

output

5

Note

In the first example subsequences [1, 2] and [4, 5] give length 4 in total.

In the second example subsequences [62, 48, 49] and [60, 61] give length 5 in total. If you choose subsequence [62, 61] in the first place then the second melody will have maximum length 2, that gives the result of 4, which is not maximal.

思路见代码注释。

#include <iostream>
#include <string>
#include <algorithm>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <queue>
#include <set>
#include <map>
#include <list>
#include <vector>
#include <stack>
#define mp make_pair
//#define P make_pair
#define MIN(a,b) (a>b?b:a)
//#define MAX(a,b) (a>b?a:b)
typedef long long ll;
typedef unsigned long long ull;
const int MAX=5e3+5;
const int MAX_V=1e3+5;
const ll INF=4e18+5;
const double M=4e18;
using namespace std;
const int MOD=1e9+7;
typedef pair<ll,int> pii;
const double eps=0.000000001;
#define rank rankk
int dp[MAX][MAX];//dp[i][j]表示第一个序列中最后一个取i，另一个序列最后一个取j的最大长度
int maxmod[10],maxnum[(int)(1e5+5)];//分别记录i一定，进行到当前j时，模7为若干或a[x]=多少时最大的dp[i][?]
int a[MAX];
int n,an;
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=0;i<=n;i++)//枚举第一个序列的情况
    {
        memset(maxmod,0,sizeof(maxmod));
        memset(maxnum,0,sizeof(maxnum));
        for(int j=1;j<=n;j++)
        {
            if(i==j)
                dp[i][j]=0;//i=j构不成两个非空不重合的序列，故为0
            else
            {
                int mod=a[j]%7;
                if(i>j)//i>j的情况转化为已算过的j<i
                {
                    maxmod[mod]=max(maxmod[mod],dp[i][j]);
                    maxnum[a[j]]=max(maxnum[a[j]],dp[i][j]);
                    continue;
                }
                dp[i][j]=max(maxmod[mod]+1,maxnum[a[j]-1]+1);//前一个为mod7同余 或 前一个为a[j]-1 的比较
                dp[i][j]=max(dp[i][j],maxnum[a[j]+1]+1);//与前一个为a[i]+1的比较
                dp[i][j]=max(dp[i][j],dp[i][0]+1);//新起一个序列
                dp[j][i]=dp[i][j];
                maxmod[mod]=max(maxmod[mod],dp[i][j]);
                maxnum[a[j]]=max(maxnum[a[j]],dp[i][j]);
                an=max(an,dp[i][j]);
            }
        }
    }
    printf("%d
",an);
    return 0;
}