Codeforces-813D Two Melodies(dp)

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

41 2 4 5

output

4

input

662 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.

设dp[i][j]为2个子序列的结尾分别为i和j时的最大总产度，可以令i<j并枚举i更新j的结尾，这样就可以避免重复：如果i>j并枚举i，则当前的j有可能是已经被第一个子序列选择过的。由于dp[i][j]和dp[j][i]是相等的，因此每次更新完dp[i][j]可以令dp[j][i]=dp[i][j]，这样自然就求出了i>j时的情况

#include <bits/stdc++.h>#define x first#define y secondusing namespace std;typedef long long LL;const int MX = 5005;const int MXM = 1e5 + 5;int dp[MX][MX]; //dp[i][j]表示以i，j结尾的2个子序列的总长度int a[MX], pre[MXM], mod[8];int main() {    //freopen("in.txt", "r", stdin);    int n, ans = 0;    scanf("%d", &n);    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);    for (int i = 0; i <= n; i++) {  //枚举结尾小的i，更新结尾大的j，从而保证不重复        memset(pre, 0, sizeof(pre));        memset(mod, 0, sizeof(mod));        for (int j = 1; j < i; j++) {            pre[a[j]] = max(pre[a[j]], dp[i][j]);//pre维护前缀中结尾为a[j]的最大总长度            mod[a[j] % 7] = max(mod[a[j] % 7], dp[i][j]);//mod维护前缀中结尾模7后的最大总长度        }        for (int j = i + 1; j <= n; j++) {            dp[i][j] = max(pre[a[j] + 1], pre[a[j] - 1]) + 1;            dp[i][j] = max(dp[i][j], mod[a[j] % 7] + 1);            dp[i][j] = max(dp[i][j], dp[i][0] + 1);            dp[j][i] = dp[i][j];    //对称            pre[a[j]] = max(pre[a[j]], dp[i][j]);            mod[a[j] % 7] = max(mod[a[j] % 7], dp[i][j]);            ans = max(ans, dp[i][j]);        }    }    printf("%d\n", ans);    return 0;}