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 <typename T, typename C = T>
struct min_cost_flow {
  struct edge {
    int src, dst;
    T cap, flow;
    C cost;
    int rev;
  };
  int n;
  vector<vector<edge>> adj;
  min_cost_flow(int n) : n(n), adj(n) {}
  void add_edge(int src, int dst, T cap, C cost) {
    adj[src].push_back({src, dst, cap, 0, cost, (int)adj[dst].size()});
    if (src == dst) adj[src].back().rev++;
    adj[dst].push_back({dst, src, 0, 0, -cost, (int)adj[src].size() - 1});
  }
  const C oo = numeric_limits<C>::max();
  vector<C> dist, pot;
  vector<edge *> prev;
  bool dijkstra(int s, int t) {
    dist.assign(n, oo);
    prev.assign(n, nullptr);
    dist[s] = 0;
    using pci = pair<C, int>;
    priority_queue<pci, vector<pci>, greater<pci>> pq;
    pq.push({0, s});
    while (!pq.empty()) {
      C d;
      int u;
      tie(d, u) = pq.top();
      pq.pop();
      if (d != dist[u]) continue;
      for (auto &e : adj[u])
        if (e.flow < e.cap &&
            dist[e.dst] > dist[u] + e.cost + pot[u] - pot[e.dst]) {
          dist[e.dst] = dist[u] + e.cost + pot[u] - pot[e.dst];
          prev[e.dst] = &e;
          pq.push({dist[e.dst], e.dst});
        }
    }
    return dist[t] < oo;
  }
  pair<T, C> max_flow(int s, int t) {
    T flow = 0;
    C cost = 0;
    pot.assign(n, 0);
    while (dijkstra(s, t)) {
      for (int u = 0; u < n; ++u)
        if (dist[u] < oo) pot[u] += dist[u];
      T delta = numeric_limits<T>::max();
      for (edge *e = prev[t]; e != nullptr; e = prev[e->src])
        delta = min(delta, e->cap - e->flow);
      flow += delta;
      for (edge *e = prev[t]; e != nullptr; e = prev[e->src]) {
        e->flow += delta;
        adj[e->dst][e->rev].flow -= delta;
        cost += delta * e->cost;
      }
    }
    return {flow, cost};
  }
};
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (auto &i : a) cin >> i;
  min_cost_flow<int, int> mcf(2 * n + 3);
  for (int i = 0; i < n; ++i) {
    mcf.add_edge(2 * n + 1, i + 1, 1, 0);
    mcf.add_edge(n + i + 1, 2 * n + 2, 1, 0);
    mcf.add_edge(i + 1, n + i + 1, 1, -1);
    for (int j = i + 1; j < n; ++j)
      if (abs(a[i] - a[j]) == 1 || a[i] % 7 == a[j] % 7)
        mcf.add_edge(n + i + 1, j + 1, 1, 0);
  }
  mcf.add_edge(0, 2 * n + 1, 4, 0);
  cout << -mcf.max_flow(0, 2 * n + 2).second << "\n";
  return 0;
}