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;
template <class T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
  os << '{';
  for (int i = (0), __i = (((int)(vec).size())); i < __i; i++) {
    os << vec[i];
    if (i + 1 != ((int)(vec).size())) os << ',';
  }
  os << '}';
  return os;
}
template <class T1, class T2>
ostream &operator<<(ostream &os, const pair<T1, T2> &par) {
  os << '(' << par.first << ',' << par.second << ')';
  return os;
}
int gcd(int x, int y) { return y ? gcd(y, x % y) : abs(x); }
template <class T>
T sqr(T x) {
  return x * x;
}
int a[5001];
int dp[5001][5001];
int maxmod[7];
int maxnum[100002];
int main() {
  ios::sync_with_stdio(false);
  int n;
  cin >> n;
  for (int i = (0), __i = (n); i < __i; i++) cin >> a[i + 1];
  int res = 0;
  for (int y = (0), __y = (n); y < __y; y++) {
    for (int i = (0), __i = (100001); i < __i; i++) maxnum[i] = dp[y][0];
    for (int i = (0), __i = (7); i < __i; i++) maxmod[i] = dp[y][0];
    for (int j = (1), __j = (y); j < __j; j++) {
      maxnum[a[j]] = max(dp[y][j], maxnum[a[j]]);
      maxmod[a[j] % 7] = max(dp[y][j], maxmod[a[j] % 7]);
    }
    for (int i = (y + 1), __i = (n + 1); i < __i; i++) {
      dp[i][y] =
          max(max(maxnum[a[i] - 1], maxnum[a[i] + 1]), maxmod[a[i] % 7]) + 1;
      maxnum[a[i]] = max(dp[i][y], maxnum[a[i]]);
      maxmod[a[i] % 7] = max(dp[i][y], maxmod[a[i] % 7]);
      res = max(res, dp[i][y]);
    }
  }
  cout << res;
  return 0;
}