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;
template <class T>
ostream& operator<<(ostream& os, vector<T> V) {
  os << "[ ";
  for (auto v : V) os << v << " ";
  return os << "]";
}
template <class L, class R>
ostream& operator<<(ostream& os, pair<L, R> P) {
  return os << "(" << P.first << "," << P.second << ")";
}
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  cout << name << " : " << arg1 << std::endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  cout.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const long long Mod = 1e9 + 7;
const int N = 5e3 + 7;
const int V = 1e5 + 7;
void update(int& x, int y) {
  if (x < y) x = y;
}
int dp[N][N], bestMod[7], bestValue[V], A[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> A[i];
  int ans = 0;
  for (int j = 0; j <= n; j++) {
    memset(bestMod, 0, sizeof(bestMod));
    for (int i = 1; i <= n; i++) bestValue[A[i]] = 0;
    for (int i = 1; i <= j; i++) {
      update(bestMod[A[i] % 7], dp[i][j]);
      update(bestValue[A[i]], dp[i][j]);
    }
    for (int i = j + 1; i <= n; i++) {
      update(dp[i][j], bestMod[A[i] % 7] + 1);
      update(dp[i][j], bestValue[A[i] - 1] + 1);
      update(dp[i][j], bestValue[A[i] + 1] + 1);
      update(dp[i][j], 1 + dp[0][j]);
      update(bestMod[A[i] % 7], dp[i][j]);
      update(bestValue[A[i]], dp[i][j]);
      update(ans, dp[i][j]);
      dp[j][i] = dp[i][j];
    }
  }
  cout << ans << endl;
  return 0;
}