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).
#include <bits/stdc++.h>
using namespace std;
const int INF = 1 << 30;
struct Primal_Dual {
  struct edge {
    int to, cap, cost, rev;
  };
  vector<vector<edge> > graph;
  vector<int> potential, min_cost, prevv, preve;
  Primal_Dual(int V) : graph(V) {}
  void add_edge(int from, int to, int cap, int cost) {
    graph[from].push_back((edge){to, cap, cost, (int)graph[to].size()});
    graph[to].push_back((edge){from, 0, -cost, (int)graph[from].size() - 1});
  }
  int min_cost_flow(int s, int t, int f) {
    int V = graph.size(), ret = 0;
    priority_queue<pair<int, int>, vector<pair<int, int> >,
                   greater<pair<int, int> > >
        que;
    potential.assign(V, 0);
    preve.assign(V, -1);
    prevv.assign(V, -1);
    while (f > 0) {
      min_cost.assign(V, INF);
      que.push(pair<int, int>(0, s));
      min_cost[s] = 0;
      while (!que.empty()) {
        pair<int, int> p = que.top();
        que.pop();
        if (min_cost[p.second] < p.first) continue;
        for (int i = 0; i < graph[p.second].size(); i++) {
          edge &e = graph[p.second][i];
          int nextCost = min_cost[p.second] + e.cost + potential[p.second] -
                         potential[e.to];
          if (e.cap > 0 && min_cost[e.to] > nextCost) {
            min_cost[e.to] = nextCost;
            prevv[e.to] = p.second, preve[e.to] = i;
            que.push(pair<int, int>(min_cost[e.to], e.to));
          }
        }
      }
      if (min_cost[t] == INF) return -1;
      for (int v = 0; v < V; v++) potential[v] += min_cost[v];
      int addflow = f;
      for (int v = t; v != s; v = prevv[v]) {
        addflow = min(addflow, graph[prevv[v]][preve[v]].cap);
      }
      f -= addflow;
      ret += addflow * potential[t];
      for (int v = t; v != s; v = prevv[v]) {
        edge &e = graph[prevv[v]][preve[v]];
        e.cap -= addflow;
        graph[v][e.rev].cap += addflow;
      }
    }
    return ret;
  }
};
int main() {
  int N, A[5000];
  cin >> N;
  for (int i = 0; i < N; i++) {
    cin >> A[i];
  }
  deque<int> latte[7], malta[100002];
  for (int i = 0; i < N; i++) {
    latte[A[i] % 7].push_back(i);
    malta[A[i]].push_back(i);
  }
  Primal_Dual flow(N * 2 + 2);
  for (int i = 0; i < N; i++) {
    latte[A[i] % 7].pop_front();
    malta[A[i]].pop_front();
    for (int j = 0; j < 300; j++) {
      if (j < latte[A[i] % 7].size())
        flow.add_edge(i, latte[A[i] % 7][j] + N, 1, 0);
      if (j < malta[A[i] - 1].size())
        flow.add_edge(i, malta[A[i] - 1][j] + N, 1, 0);
      if (j < malta[A[i] + 1].size())
        flow.add_edge(i, malta[A[i] + 1][j] + N, 1, 0);
    }
    flow.add_edge(i + N, i, 1, -1);
    flow.add_edge(i, N * 2 + 1, 1, 0);
    flow.add_edge(N * 2, i + N, 1, 0);
  }
  cout << -flow.min_cost_flow(N * 2, N * 2 + 1, 4) << endl;
}
