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;
const int maxn = 5e3 + 55;
int dp[maxn][maxn];
int max_mod[11], maxnum[(int)(1e5 + 5)];
int arr[maxn];
int n, ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);
  for (int i = 0; i <= n; i++) {
    memset(max_mod, 0, sizeof(max_mod));
    memset(maxnum, 0, sizeof(maxnum));
    for (int j = 1; j <= n; j++) {
      if (i == j)
        dp[i][j] = 0;
      else {
        int md = arr[j] % 7;
        if (i > j) {
          max_mod[md] = max(max_mod[md], dp[i][j]);
          maxnum[arr[j]] = max(maxnum[arr[j]], dp[i][j]);
          continue;
        }
        dp[i][j] = max(max_mod[md] + 1, maxnum[arr[j] - 1] + 1);
        dp[i][j] = max(dp[i][j], maxnum[arr[j] + 1] + 1);
        dp[i][j] = max(dp[i][j], dp[i][0] + 1);
        dp[j][i] = dp[i][j];
        max_mod[md] = max(max_mod[md], dp[i][j]);
        maxnum[arr[j]] = max(maxnum[arr[j]], dp[i][j]);
        ans = max(ans, dp[i][j]);
      }
    }
  }
  printf("%d\n", ans);
  return 0;
}