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 N = 5005;
int n, a[N], M[100005], Max[7][N], dp[N][N], ycl[N], b[N], sum, ans;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    if (M[a[i]])
      b[i] = M[a[i]];
    else {
      M[a[i]] = b[i] = ++sum;
      ycl[sum] = a[i] % 7;
    }
  }
  for (int i = 1; i <= n; i++) {
    int zs = a[i] % 7;
    for (int j = 1; j <= sum; j++)
      Max[zs][j] =
          max(dp[b[i]][j] =
                  max(Max[zs][j], max(dp[b[i]][j], max(dp[M[a[i] + 1]][j],
                                                       dp[M[a[i] - 1]][j]))) +
                  1,
              Max[zs][j]);
    for (int j = 1; j <= sum; j++)
      Max[ycl[j]][b[i]] = max(Max[ycl[j]][b[i]], dp[j][b[i]] = dp[b[i]][j]);
  }
  for (int i = 0; i < 7; i++)
    for (int j = 1; j <= sum; j++) ans = max(ans, Max[i][j]);
  cout << ans << endl;
}
