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 = 6030;
struct MCMF {
  using T = int;
  static const T inf = 1e9 + 7;
  int n, m, s, t;
  bool vis[N];
  int par[N], pos[N];
  T pot[N], dis[N], left[N];
  priority_queue<pair<T, int>> q;
  struct edge {
    int to, rev;
    T cap, cost, flow;
    int id;
  };
  vector<edge> ed[N];
  MCMF() {}
  MCMF(int _n) {
    n = _n + 2;
    for (int i = 1; i <= n; i++) ed[i].clear();
  }
  void add_edge(int u, int v, T cap, T cost, int id = -1) {
    edge a = {v, (int)ed[v].size(), cap, cost, 0, id};
    edge b = {u, (int)ed[u].size(), 0, -cost, 0, -1};
    ed[u].push_back(a);
    ed[v].push_back(b);
  }
  T BellmanFord(int u) {
    if (vis[u]) return pot[u];
    if (u == s) {
      pot[s] = 0;
      return 0;
    }
    vis[u] = true;
    pot[u] = inf;
    for (edge e : ed[u]) {
      edge r = ed[e.to][e.rev];
      if (r.flow < r.cap) pot[u] = min(pot[u], BellmanFord(e.to) + r.cost);
    }
    return pot[u];
  }
  bool dijkstra() {
    memset(vis, 0, sizeof vis);
    for (int i = 1; i <= n; i++) dis[i] = left[i] = inf;
    dis[s] = 0;
    q.push({0, s});
    int u, v;
    while (!q.empty()) {
      u = q.top().second;
      q.pop();
      if (vis[u]) continue;
      vis[u] = true;
      int ptr = 0;
      for (edge e : ed[u]) {
        v = e.to;
        T cost = e.cost + pot[u] - pot[v];
        if (e.flow < e.cap && dis[u] + cost < dis[v]) {
          dis[v] = dis[u] + cost;
          par[v] = u;
          pos[v] = ptr;
          left[v] = min(left[u], e.cap - e.flow);
          q.push(make_pair(-dis[v], v));
        }
        ++ptr;
      }
    }
    for (int i = 1; i <= n; i++) dis[i] += (pot[i] - pot[s]);
    return vis[t];
  }
  pair<T, T> solve(int _s, int _t, T goal = inf) {
    s = _s;
    t = _t;
    memset(pot, 0, sizeof pot);
    memset(vis, 0, sizeof vis);
    BellmanFord(t);
    int u, v;
    T ans = 0, cost = 0, f;
    while (ans < goal && dijkstra()) {
      u = t;
      f = left[t];
      while (u != s) {
        v = par[u];
        ed[v][pos[u]].flow += left[t];
        ed[u][ed[v][pos[u]].rev].flow -= left[t];
        u = v;
      }
      T need = goal - ans;
      f = min(f, need);
      ans += f;
      cost += f * dis[t];
      memcpy(pot, dis, sizeof dis);
    }
    return make_pair(ans, cost);
  }
};
int a[N];
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  MCMF F(n * 2 + 4);
  int s = 2 * n + 1, t = s + 1;
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (abs(a[i] - a[j]) == 1 || abs(a[i] - a[j]) % 7 == 0) {
        F.add_edge(i + n, j, 1, 0);
      }
    }
    F.add_edge(s, i, 1, 0);
    F.add_edge(i + n, t, 1, 0);
    F.add_edge(i, i + n, 1, -1);
  }
  cout << -F.solve(s, t, 4).second << '\n';
  return 0;
}