Problem: Author note: I think some of you might remember the problem "Two Melodies" from Eductational Codeforces Round 22. Now it's time to make it a bit more difficult!

Alice is a composer, and recently she had recorded two tracks that became very popular. Now she has got a lot of fans who are waiting for new tracks. 

This time Alice wants to form four melodies for her tracks.

Alice has a sheet with n notes written on it. She wants to take four such non-empty non-intersecting subsequences that all 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 differ by 1 or are congruent modulo 7.

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

Input

The first line contains one integer number n (4 ≤ n ≤ 3000).

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 four non-empty non-intersecting subsequences that all of them form a melody.

Examples

Input

5
1 3 5 7 9


Output

4


Input

5
1 3 5 7 2


Output

5

Note

In the first example it is possible to compose 4 melodies by choosing any 4 notes (and each melody will consist of only one note).

In the second example it is possible to compose one melody with 2 notes — {1, 2}. Remaining notes are used in other three melodies (one note per each melody).

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 100000 + 10;
const int M = 1000000007;
const double PI = atan(1) * 4;
const int oo = 1000000000;
int n, dp[5001][5001], v[5001], mxm[7], mxn[N];
int main() {
  cin >> n;
  for (int i = 1; i < n + 1; ++i) scanf("%d", &v[i]);
  int ans = 0;
  for (int j = 0; j <= n; ++j) {
    memset(mxn, 0, sizeof(mxn));
    memset(mxm, 0, sizeof(mxm));
    for (int i = 0; i <= n; ++i) {
      dp[i][j] = dp[j][i];
      int d = v[i];
      if (i > j) {
        dp[i][j] = 1 + mxm[d % 7];
        dp[i][j] = max(dp[i][j], 1 + dp[0][j]);
        dp[i][j] = max(dp[i][j], 1 + mxn[d - 1]);
        dp[i][j] = max(dp[i][j], 1 + mxn[d + 1]);
      }
      mxm[d % 7] = max(mxm[d % 7], dp[i][j]);
      mxn[d] = max(mxn[d], dp[i][j]);
      ans = max(ans, dp[i][j]);
    }
  }
  cout << ans << endl;
}