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 best_val[100111], best_mod[10];
int dp[5111][5111], arr[5111];
int main() {
  int n, ans = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", arr + i);
  dp[0][0] = 0;
  for (int j = 0; j <= n; j++) {
    for (int i = 1; i <= n; i++) best_val[arr[i]] = 0;
    memset(best_mod, 0, sizeof(best_mod));
    for (int i = 0; i <= n; i++) {
      if (i < j) dp[i][j] = dp[j][i];
      if (i == j) dp[i][j] = 0;
      if (i > j) {
        int max_val = 0;
        max_val = max(max_val, best_val[arr[i] - 1]);
        max_val = max(max_val, best_val[arr[i] + 1]);
        max_val = max(max_val, best_mod[arr[i] % 7]);
        max_val = max(max_val, dp[j][0]);
        dp[i][j] = max_val + 1;
        ans = max(ans, dp[i][j]);
      }
      if (i > 0) {
        best_val[arr[i]] = max(best_val[arr[i]], dp[i][j]);
        best_mod[arr[i] % 7] = max(best_mod[arr[i] % 7], dp[i][j]);
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}
