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;
void update_max(short &x, int value) { x = max<short>(x, value); }
short dp[5000][5000] = {0};
short best[5000][5000] = {0};
short best_fwd[5000][5000] = {0};
short best_mod[5000][7] = {0};
short best_mod_fwd[5000][7] = {0};
int main() {
  int n;
  cin >> n;
  set<int> st;
  vector<int> data(n);
  for (long long _n = n, i = 0; i < _n; ++i) {
    cin >> data[i];
    st.insert(data[i]);
  }
  int s = st.size();
  vector<short> data_to_position(100002, -1);
  int index = 0;
  for (auto it : st) {
    data_to_position[it] = index;
    ++index;
  }
  vector<short> single(n);
  short result = 0;
  for (long long _n = n, i = 0; i < _n; ++i) {
    single[i] = 1;
    for (long long _n = i, j = 0; j < _n; ++j) {
      if ((data[i] - data[j]) % 7 == 0 || abs(data[i] - data[j]) == 1) {
        update_max(single[i], single[j] + 1);
      }
    }
    for (long long _n = i, j = 0; j < _n; ++j) {
      dp[i][j] = single[j] + 1;
      short pos_i = data_to_position[data[i]];
      short pos_i_plus = data_to_position[data[i] + 1];
      short pos_i_minus = data_to_position[data[i] - 1];
      short pos_j = data_to_position[data[j]];
      short pos_j_plus = data_to_position[data[j] + 1];
      short pos_j_minus = data_to_position[data[j] - 1];
      update_max(dp[i][j], 1 + best_mod[j][data[i] % 7]);
      if (pos_i_plus != -1) update_max(dp[i][j], 1 + best[j][pos_i_plus]);
      if (pos_i_minus != -1) update_max(dp[i][j], 1 + best[j][pos_i_minus]);
      update_max(dp[i][j], 1 + best_mod_fwd[j][data[i] % 7]);
      if (pos_i_plus != -1) update_max(dp[i][j], 1 + best_fwd[j][pos_i_plus]);
      if (pos_i_minus != -1) update_max(dp[i][j], 1 + best_fwd[j][pos_i_minus]);
      update_max(best_mod[i][data[j] % 7], dp[i][j]);
      update_max(best[i][pos_j], dp[i][j]);
      update_max(best_mod_fwd[j][data[i] % 7], dp[i][j]);
      update_max(best_fwd[j][pos_i], dp[i][j]);
      update_max(result, dp[i][j]);
    }
  }
  cout << result;
}