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 N = 5e3 + 1;
int n, a[N], b[N][N], c[N][7], d[N];
vector<pair<int, int> > v;
vector<int> v1;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    v.push_back({a[i], i});
  }
  sort(v.begin(), v.end());
  v1.push_back(v[0].first);
  for (int i = 0; i < (int)v.size(); i++) {
    if (v[i].first != v1.back()) v1.push_back(v[i].first);
    a[v[i].second] = v1.size() - 1;
  }
  int ans = 0, tmp, pos;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < i; j++) {
      pos = a[i];
      tmp = (j == 0) ? 1 : 2;
      tmp = max(tmp, d[j] + 1);
      tmp = max(tmp, c[j][v1[pos] % 7] + 1);
      if (pos - 1 >= 0 && v1[pos] - v1[pos - 1] == 1)
        tmp = max(tmp, b[j][pos - 1] + 1);
      if (pos + 1 < v1.size() && v1[pos + 1] - v1[pos] == 1)
        tmp = max(tmp, b[j][pos + 1] + 1);
      ans = max(ans, tmp);
      b[j][pos] = max(b[j][pos], tmp);
      c[j][v1[pos] % 7] = max(c[j][v1[pos] % 7], tmp);
      if (j > 0) {
        b[i][a[j]] = max(b[i][a[j]], tmp);
        c[i][v1[a[j]] % 7] = max(c[i][v1[a[j]] % 7], tmp);
      } else
        d[i] = max(d[i], tmp);
    }
  }
  cout << ans << '\n';
}