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>
using namespace std;
template <class T>
inline void cmax(T& x, int y) {
  if (x < y) x = y;
}
int main() {
  int n, ans = 0;
  scanf("%d", &n);
  vector<int> a(n + 1, 0);
  vector<short> d(n + 1, 0);
  vector<vector<short>> m(n + 1, vector<short>(3, 0));
  vector<vector<short>> p(n + 1, vector<short>(7, 0));
  vector<vector<short>> q(n + 1, vector<short>(n * 3, 0));
  map<int, int> M;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    for (int j = a[i] - 1; j <= a[i] + 1; j++) {
      if (M.find(j) == M.end()) {
        int t = M.size();
        M[j] = t;
      }
      m[i][j - a[i] + 1] = M[j];
    }
    a[i] %= 7;
    for (int j = i - 1; j >= 0; j--) {
      int tmp = p[j][a[i]];
      cmax(tmp, q[j][m[i][0]]);
      cmax(tmp, q[j][m[i][2]]);
      cmax(ans, ++tmp);
      cmax(p[j][a[i]], tmp);
      cmax(q[j][m[i][1]], tmp);
      cmax(p[i][a[j]], tmp);
      cmax(q[i][m[j][1]], tmp);
      if (j == 0)
        for (int k = 0; k < 7; k++) cmax(p[i][k], tmp);
    }
  }
  printf("%d\n", ans);
}
