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 = 6005, MX = 1E5 + 5, INF = 1E8 + 7;
int n, u, ind = 0, pot[N], dst[N], lv[MX], lm[7];
int s, t, ans = 0, pa[MX], we[MX], co[MX], tr[MX];
vector<pair<int, int>> adj[N];
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>>
    pq;
void add_edge(int u, int v, int w, int c) {
  we[ind] = w;
  co[ind] = c;
  pa[ind] = u ^ v;
  adj[u].push_back({v, ind++});
  we[ind] = 0;
  co[ind] = -c;
  pa[ind] = u ^ v;
  adj[v].push_back({u, ind++});
  pot[v] = max(pot[v], pot[u] + c);
}
void dijkstra() {
  fill(dst + 1, dst + t + 1, INF);
  pq.push({dst[s] = 0, 0});
  while (!pq.empty()) {
    pair<int, int> u = pq.top();
    pq.pop();
    if (u.first > dst[u.second]) {
      continue;
    }
    for (pair<int, int> &v : adj[u.second]) {
      int cur = u.first + co[v.second] + pot[u.second] - pot[v.first];
      if (cur < dst[v.first] && we[v.second] > 0) {
        tr[v.first] = v.second;
        pq.push({dst[v.first] = cur, v.first});
      }
    }
  }
}
void trace() {
  int u = t, mx = INF;
  while (u != s) {
    int ind = tr[u];
    mx = min(mx, we[ind]);
    u = u ^ pa[ind];
  }
  u = t;
  while (u != s) {
    int ind = tr[u];
    we[ind] -= mx;
    we[ind ^ 1] += mx;
    ans += mx * co[ind];
    u = u ^ pa[ind];
  }
  for (int i = 0; i <= t; i++) {
    pot[i] += dst[i];
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  s = 0, t = 2 * n + 1;
  fill(pot + 1, pot + t + 1, -INF);
  for (int i = 1; i <= n; i++) {
    cin >> u;
    if (lm[u % 7] != 0) {
      add_edge(lm[u % 7] + n, i, 1, -1);
      add_edge(lm[u % 7], i, INF, 0);
    }
    lm[u % 7] = i;
    if (lv[u - 1] != 0) {
      add_edge(lv[u - 1] + n, i, 1, -1);
    }
    if (lv[u + 1] != 0) {
      add_edge(lv[u + 1] + n, i, 1, -1);
    }
    if (lv[u] != 0) {
      add_edge(lv[u], i, INF, 0);
    }
    lv[u] = i;
    add_edge(s, i, 1, -1);
    add_edge(i, i + n, 1, 0);
    add_edge(i + n, t, 1, 0);
  }
  for (int i = 0; i < 4; i++) {
    dijkstra();
    trace();
  }
  cout << -ans;
}