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;
int n;
int a[5001];
int dp[5001][5001];
int lmod[7];
int lnum[100011];
int main() {
  int ans = 0;
  cin >> n;
  for (int i = 1; i < n + 1; i++) {
    cin >> a[i];
  }
  for (int i = 0; i < n + 1; i++) {
    memset(lmod, 0, sizeof(lmod));
    for (int j = 1; j < n + 1; j++) {
      if (j > i)
        dp[i][j] = dp[j][i] = max(max(dp[i][0], lmod[a[j] % 7]),
                                  max(lnum[a[j] - 1], lnum[a[j] + 1])) +
                              1;
      lnum[a[j]] = max(lnum[a[j]], dp[i][j]);
      lmod[a[j] % 7] = max(lmod[a[j] % 7], dp[i][j]);
      ans = max(ans, dp[i][j]);
    }
    for (int j = 1; j < n + 1; j++) lnum[a[j]] = 0;
  }
  cout << ans << endl;
}
