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;
long long dp[5005][5005];
long long a[5005];
long long n;
long long i, j, k;
long long ans;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  ans = 0;
  for (int y = 0; y <= n; y++) {
    long long maxmod[100009];
    long long maxnum[100009];
    memset(maxmod, 0, sizeof(maxmod));
    memset(maxnum, 0, sizeof(maxnum));
    for (int x = 0; x < y; x++) {
      maxmod[a[x] % 7] = max(maxmod[a[x] % 7], dp[y][x]);
      maxnum[a[x]] = max(maxnum[a[x]], dp[y][x]);
    }
    for (int x = y + 1; x <= n; x++) {
      dp[x][y] = dp[y][x] = max(max(maxmod[a[x] % 7], dp[y][0]),
                                max(maxnum[a[x] - 1], maxnum[a[x] + 1])) +
                            1;
      ans = max(ans, dp[x][y]);
      maxmod[a[x] % 7] = max(maxmod[a[x] % 7], dp[x][y]);
      maxnum[a[x]] = max(maxnum[a[x]], dp[x][y]);
    }
  }
  cout << ans << endl;
}
