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;
void abc() { cerr << endl; }
template <typename T, typename... U>
void abc(T a, U... b) {
  cerr << a << ' ', abc(b...);
}
template <typename T>
void printv(T l, T r) {
  while (l != r) cout << *l << " \n"[++l == r];
}
template <typename A, typename B>
istream& operator>>(istream& o, pair<A, B>& a) {
  return o >> a.first >> a.second;
}
template <typename A, typename B>
ostream& operator<<(ostream& o, pair<A, B> a) {
  return o << '(' << a.first << ", " << a.second << ')';
}
template <typename T>
ostream& operator<<(ostream& o, vector<T> a) {
  bool is = false;
  for (T i : a) {
    o << (is ? ' ' : '{'), is = true, o << i;
  }
  return o << '}';
}
template <typename T>
struct vv : vector<vector<T>> {
  vv(int n, int m, T v) : vector<vector<T>>(n, vector<T>(m, v)) {}
  vv() {}
};
template <typename T>
struct vvv : vector<vv<T>> {
  vvv(int n, int m, int k, T v) : vector<vv<T>>(n, vv<T>(m, k, v)) {}
  vvv() {}
};
const int mod = 1e9 + 7, N = 5001, logN = 20, K = 111;
struct MCMF {
  const int INF = 1 << 30;
  struct edge {
    int v, id, revid, f, c;
    edge(int _v, int _f, int _c, int _id, int _revid)
        : v(_v), f(_f), c(_c), id(_id), revid(_revid) {}
  };
  vector<vector<edge>> adj;
  vector<pair<int, int>> rt;
  vector<int> dis;
  int n, s, t;
  MCMF(int _n, int _s, int _t) : n(_n), s(_s), t(_t) { adj.resize(n); }
  void add_edge(int u, int v, long long f, long long c) {
    void(0);
    adj[u].push_back(edge(v, f, c, adj[u].size(), adj[v].size()));
    adj[v].push_back(edge(u, 0, -c, adj[v].size(), adj[u].size() - 1));
  }
  bool SPFA() {
    rt.assign(n, make_pair(-1, -1));
    dis.assign(n, INF);
    vector<bool> vis(n, false);
    queue<int> q;
    q.push(s);
    dis[s] = 0;
    vis[s] = true;
    while (q.size()) {
      int v = q.front();
      q.pop();
      vis[v] = false;
      for (edge& e : adj[v])
        if (e.f > 0 && dis[e.v] > dis[v] + e.c) {
          dis[e.v] = dis[v] + e.c;
          rt[e.v] = make_pair(v, e.id);
          if (!vis[e.v]) {
            vis[e.v] = true;
            q.push(e.v);
          }
        }
    }
    return dis[t] != INF;
  }
  pair<int, int> runFlow() {
    int cost = 0, flow = 0;
    while (SPFA()) {
      vector<pair<int, int>> E;
      int v = t;
      int addflow = INF;
      while (v != s) {
        E.push_back(rt[v]);
        addflow = min(addflow, adj[rt[v].first][rt[v].second].f);
        v = rt[v].first;
      }
      for (pair<int, int> a : E) {
        adj[a.first][a.second].f -= addflow;
        adj[adj[a.first][a.second].v][adj[a.first][a.second].revid].f +=
            addflow;
      }
      flow += addflow;
      cost += addflow * dis[t];
    }
    return make_pair(cost, flow);
  }
};
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; ++i) cin >> a[i];
  MCMF flow(n * 4 + 4, n * 4, n * 4 + 2);
  for (int i = 0; i < n; ++i) flow.add_edge(i + n, i, 1, -1);
  flow.add_edge(n * 4, n * 4 + 1, 4, 0);
  for (int i = 0; i < n; ++i) flow.add_edge(n * 4 + 1, i + n, 1, 0);
  for (int i = 0; i < n; ++i) flow.add_edge(i, n * 4 + 3, 1, 0);
  flow.add_edge(n * 4 + 3, n * 4 + 2, 4, 0);
  int ty = n * 2;
  vector<int> pre(7, -1);
  for (int i = 0; i < n; ++i) {
    int r = a[i] % 7;
    if (pre[r] != -1) {
      flow.add_edge(pre[r], ty, 1 << 20, 0);
    }
    flow.add_edge(i, ty, 1, 0);
    if (pre[r] != -1) flow.add_edge(pre[r], i + n, 1, 0);
    pre[r] = ty++;
  }
  pre.assign(100002, -1);
  for (int i = 0; i < n; ++i) {
    if (pre[a[i]] != -1) {
      flow.add_edge(pre[a[i]], ty, 1 << 20, 0);
    }
    flow.add_edge(i, ty, 1, 0);
    if (pre[a[i] - 1] != -1) flow.add_edge(pre[a[i] - 1], i + n, 1, 0);
    if (pre[a[i] + 1] != -1) flow.add_edge(pre[a[i] + 1], i + n, 1, 0);
    pre[a[i]] = ty++;
  }
  pair<int, int> ans = flow.runFlow();
  assert(ans.second == 4);
  cout << -ans.first << endl;
}
