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;
const int maxn = 100010;
const int INF = 1e9 + 10;
int n;
int a[maxn];
int dp[5010][5010];
int maxmod[7], maxnum[maxn];
int main() {
  while (cin >> n) {
    int MaxA = 0;
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), MaxA = max(MaxA, a[i]);
    for (int y = 0; y <= n; y++) {
      memset(maxmod, 0, sizeof(maxmod));
      for (int i = 1; i <= n; i++) maxnum[a[i]] = 0;
      maxnum[0] = maxnum[MaxA + 1] = 0;
      for (int x = 0; x <= n; x++) {
        if (x == y)
          dp[x][y] = 0;
        else if (x < y)
          dp[x][y] = dp[y][x];
        else {
          dp[x][y] = max(max(maxmod[a[x] % 7] + 1, dp[0][y] + 1),
                         max(maxnum[a[x] - 1] + 1, maxnum[a[x] + 1] + 1));
        }
        maxnum[a[x]] = max(maxnum[a[x]], dp[x][y]);
        maxmod[a[x] % 7] = max(maxmod[a[x] % 7], dp[x][y]);
      }
    }
    int res = 0;
    for (int i = 1; i <= n; i++)
      for (int j = 1; j <= n; j++) res = max(res, dp[i][j]);
    cout << res << endl;
  }
  return 0;
}
