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).
#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
function<void(void)> ____ = []() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
};
const int MAXN = 5555;
void solve() {
  int n;
  scanf("%d", &n);
  vector<int> A(n + 1);
  for (int i = 1; i <= n; i++) scanf("%d", &A[i]);
  vector<vector<int> > f(n + 1, vector<int>(n + 1));
  vector<int> maxmod(7, 0), maxnum(100007, 0);
  for (int i = 0; i <= n; i++) {
    for (int j = 1; j < i; j++)
      ((maxmod[A[j] % 7]) =
           (maxmod[A[j] % 7]) > (f[i][j]) ? (maxmod[A[j] % 7]) : (f[i][j])),
          ((maxnum[A[j]]) =
               (maxnum[A[j]]) > (f[i][j]) ? (maxnum[A[j]]) : (f[i][j]));
    for (int j = i + 1; j <= n; j++) {
      f[j][i] = f[i][j] =
          max({f[i][0], maxmod[A[j] % 7], maxnum[A[j] - 1], maxnum[A[j] + 1]}) +
          1;
      ((maxmod[A[j] % 7]) =
           (maxmod[A[j] % 7]) > (f[i][j]) ? (maxmod[A[j] % 7]) : (f[i][j]));
      ((maxnum[A[j]]) =
           (maxnum[A[j]]) > (f[i][j]) ? (maxnum[A[j]]) : (f[i][j]));
    }
    fill(maxmod.begin(), maxmod.end(), 0);
    for (int j = 1; j <= n; j++) maxnum[A[j]] = 0;
  }
  int ret = 0;
  for (int i = 0; i <= n; i++)
    ((ret) = (ret) > (*max_element(f[i].begin(), f[i].end()))
                 ? (ret)
                 : (*max_element(f[i].begin(), f[i].end())));
  cout << ret << endl;
}
int main() {
  solve();
  return 0;
}
