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 Flow = int, class Cost = int>
struct MinCostFlow {
  const Flow INF_FLOW = 1000111000;
  const Cost INF_COST = 1000111000111000LL;
  int n, t, S, T;
  Flow totalFlow;
  Cost totalCost;
  vector<int> last, visited;
  vector<Cost> dis;
  struct Edge {
    int to;
    Flow cap;
    Cost cost;
    int next;
    Edge(int to, Flow cap, Cost cost, int next)
        : to(to), cap(cap), cost(cost), next(next) {}
  };
  vector<Edge> edges;
  MinCostFlow(int n)
      : n(n),
        t(0),
        totalFlow(0),
        totalCost(0),
        last(n, -1),
        visited(n, 0),
        dis(n, 0) {
    edges.clear();
  }
  int addEdge(int from, int to, Flow cap, Cost cost) {
    edges.push_back(Edge(to, cap, cost, last[from]));
    last[from] = t++;
    edges.push_back(Edge(from, 0, -cost, last[to]));
    last[to] = t++;
    return t - 2;
  }
  pair<Flow, Cost> minCostFlow(int _S, int _T) {
    S = _S;
    T = _T;
    SPFA();
    while (1) {
      while (1) {
        for (int i = 0; i < n; i++) visited[i] = 0;
        if (!findFlow(S, INF_FLOW)) break;
      }
      if (!modifyLabel()) break;
    }
    return make_pair(totalFlow, totalCost);
  }

 private:
  void SPFA() {
    for (int i = 0; i < n; i++) dis[i] = INF_COST;
    priority_queue<pair<Cost, int> > Q;
    Q.push(make_pair(dis[S] = 0, S));
    while (!Q.empty()) {
      int x = Q.top().second;
      Cost d = -Q.top().first;
      Q.pop();
      if (dis[x] != d) continue;
      for (int it = last[x]; it >= 0; it = edges[it].next)
        if (edges[it].cap > 0 && dis[edges[it].to] > d + edges[it].cost)
          Q.push(make_pair(-(dis[edges[it].to] = d + edges[it].cost),
                           edges[it].to));
    }
    Cost disT = dis[T];
    for (int i = 0; i < n; i++) dis[i] = disT - dis[i];
  }
  Flow findFlow(int x, Flow flow) {
    if (x == T) {
      totalCost += dis[S] * flow;
      totalFlow += flow;
      return flow;
    }
    visited[x] = 1;
    Flow now = flow;
    for (int it = last[x]; it >= 0; it = edges[it].next)
      if (edges[it].cap && !visited[edges[it].to] &&
          dis[edges[it].to] + edges[it].cost == dis[x]) {
        Flow tmp = findFlow(edges[it].to, min(now, edges[it].cap));
        edges[it].cap -= tmp;
        edges[it ^ 1].cap += tmp;
        now -= tmp;
        if (!now) break;
      }
    return flow - now;
  }
  bool modifyLabel() {
    Cost d = INF_COST;
    for (int i = 0; i < n; i++)
      if (visited[i])
        for (int it = last[i]; it >= 0; it = edges[it].next)
          if (edges[it].cap && !visited[edges[it].to])
            d = min(d, dis[edges[it].to] + edges[it].cost - dis[i]);
    if (d == INF_COST) return false;
    for (int i = 0; i < n; i++)
      if (visited[i]) dis[i] += d;
    return true;
  }
};
int main() {
  int n;
  scanf("%d", &n);
  int a[n];
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  MinCostFlow<int, long long> mcf(3 * (n + 2));
  mcf.addEdge(0, 1, 4, 0);
  mcf.addEdge(3 * (n + 1), 3 * (n + 1) + 1, 4, 0);
  for (int i = 0; i < n; i++) {
    mcf.addEdge(1, (i + 1) * 3, 1, 0);
    mcf.addEdge((i + 1) * 3, (i + 1) * 3 + 1, 1, -1);
    mcf.addEdge((i + 1) * 3 + 2, (i + 1) * 3, 1, 0);
    mcf.addEdge((i + 1) * 3 + 1, (n + 1) * 3, 1, 0);
    int u, d, l;
    u = d = l = 0;
    for (int j = i + 1; j < n; j++) {
      if (a[j] + 1 == a[i] && d < 4) {
        mcf.addEdge(3 * (i + 1) + 1, 3 * (j + 1), 1, 0);
        d++;
      } else if (a[j] - 1 == a[i] && u < 4) {
        mcf.addEdge(3 * (i + 1) + 1, 3 * (j + 1), 1, 0);
        u++;
      } else if (!((a[i] - a[j]) % 7) && !l) {
        mcf.addEdge(3 * (i + 1) + 1, 3 * (j + 1) + 2, 1, 0);
        mcf.addEdge(3 * (i + 1) + 2, 3 * (j + 1) + 2, 1, 0);
        l = 1;
      }
    }
  }
  pair<int, int> mcmf = mcf.minCostFlow(0, 3 * (n + 1) + 1);
  if (mcmf.first == 4)
    cout << -mcmf.second << endl;
  else
    cout << 0 << endl;
  return 0;
}