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 = 5005;
int a[N], n, f[N][N], nw[N * 20], mod[10], ans;
int Max(int A, int B, int C, int D) { return max(max(A, B), max(C, D)); }
signed main() {
  cin >> n;
  for (register int i = (1); i <= (n); ++i) cin >> a[i];
  for (register int i = (0); i <= (n); ++i) {
    memset(nw, 0, sizeof(nw));
    memset(mod, 0, sizeof(mod));
    for (register int j = (i - 1); j >= (1); --j)
      nw[a[j]] = max(nw[a[j]], f[j][i]),
      mod[a[j] % 7] = max(mod[a[j] % 7], f[j][i]);
    for (register int j = (i + 1); j <= (n); ++j) {
      f[i][j] = Max(nw[a[j] + 1], nw[a[j] - 1], mod[a[j] % 7], f[0][i]) + 1;
      nw[a[j]] = max(nw[a[j]], f[i][j]);
      mod[a[j] % 7] = max(mod[a[j] % 7], f[i][j]);
      ans = max(ans, f[i][j]);
    }
  }
  cout << ans << '\n';
  return 0;
}