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;
int v[5000 + 5], d[5000 + 5][5000 + 5];
int md[100000 + 5], mm[10];
void umax(int &a, int b) { a = max(a, b); }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  int i, j, z, ans = 0;
  for (i = 1; i <= n; i++) cin >> v[i];
  for (j = 0; j <= n; j++) {
    fill(begin(md), end(md), -1);
    fill(begin(mm), end(mm), -1);
    for (i = 1; i < j; i++) {
      umax(md[v[i]], d[j][i]);
      umax(mm[v[i] % 7], d[j][i]);
    }
    for (i = j + 1; i <= n; i++) {
      umax(d[i][j], 1 + md[v[i] - 1]);
      umax(d[i][j], 1 + md[v[i] + 1]);
      umax(d[i][j], 1 + mm[v[i] % 7]);
      umax(d[i][j], 1 + d[j][0]);
      umax(md[v[i]], d[i][j]);
      umax(mm[v[i] % 7], d[i][j]);
      umax(ans, d[i][j]);
    }
  }
  cout << ans;
  return 0;
}