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 trans[100009];
int dp[5555][5555], ndp[5555][5555];
int X[5555][8], Y[5555][8];
int main() {
  ios_base::sync_with_stdio(false), cout.tie(0), cin.tie(0);
  memset(trans, -1, sizeof trans);
  int n;
  cin >> n;
  vector<int> A(n), B(n);
  for (int i = 0; i < n; ++i) cin >> B[i], A[i] = B[i];
  B.push_back(0);
  sort(B.begin(), B.end());
  B.erase(unique(B.begin(), B.end()), B.end());
  int m = ((int)(B).size()), res = 0;
  for (int i = 0; i < m; ++i) trans[B[i]] = i;
  for (int i = 0; i < n; ++i) {
    vector<pair<pair<int, int>, int>> upd;
    int mod = A[i] % 7, T = trans[A[i]];
    for (int j = 0; j < m; ++j) {
      ndp[j][T] = max(ndp[j][T], dp[j][0] + 1);
      ndp[T][j] = max(ndp[T][j], dp[0][j] + 1);
      ndp[j][T] = max(ndp[j][T], X[j][mod] + 1);
      ndp[T][j] = max(ndp[T][j], Y[j][mod] + 1);
    }
    if (trans[A[i] - 1] != -1) {
      for (int j = 0; j < m; ++j)
        ndp[j][T] = max(ndp[j][T], dp[j][trans[A[i] - 1]] + 1),
        ndp[T][j] = max(ndp[T][j], dp[trans[A[i] - 1]][j] + 1);
    }
    if (trans[A[i] + 1] != -1) {
      for (int j = 0; j < m; ++j)
        ndp[j][T] = max(ndp[j][T], dp[j][trans[A[i] + 1]] + 1),
        ndp[T][j] = max(ndp[T][j], dp[trans[A[i] + 1]][j] + 1);
    }
    for (int j = 0; j < m; ++j) {
      int mod2 = B[j] % 7;
      if (dp[j][T] < ndp[j][T]) {
        dp[j][T] = ndp[j][T];
        X[j][mod] = max(X[j][mod], dp[j][T]);
        Y[T][mod2] = max(Y[T][mod2], dp[j][T]);
      }
      if (dp[T][j] < ndp[T][j]) {
        dp[T][j] = ndp[T][j];
        X[T][mod2] = max(X[T][mod2], dp[T][j]);
        Y[j][mod] = max(Y[j][mod], dp[T][j]);
      }
    }
  }
  for (int i = 0; i < m; ++i)
    for (int j = 0; j < m; ++j) res = max(res, dp[i][j]);
  cout << res << "\n";
  return 0;
}
