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;
int a[6100];
struct edge {
  int to, capacity, cost, rev;
  edge() {}
  edge(int to, int _capacity, int _cost, int _rev)
      : to(to), capacity(_capacity), cost(_cost), rev(_rev) {}
};
const int inf = 0x3f3f3f3f;
struct Min_Cost_Max_Flow {
  int V, H[6100 + 5], dis[6100 + 5], PreV[6100 + 5], PreE[6100 + 5];
  vector<edge> G[6100 + 5];
  void Init(int n) {
    V = n;
    for (int i = 0; i <= V; ++i) G[i].clear();
  }
  void add(int from, int to, int cap, int cost) {
    G[from].push_back(edge(to, cap, cost, G[to].size()));
    G[to].push_back(edge(from, 0, -cost, G[from].size() - 1));
  }
  int minCost(int s, int t, int f) {
    int flow = 0;
    int res = 0;
    fill(H, H + 1 + V, 0);
    while (f) {
      priority_queue<pair<int, int>, vector<pair<int, int> >,
                     greater<pair<int, int> > >
          q;
      fill(dis, dis + 1 + V, inf);
      dis[s] = 0;
      q.push(pair<int, int>(0, s));
      while (!q.empty()) {
        pair<int, int> now = q.top();
        q.pop();
        int v = now.second;
        if (dis[v] < now.first) continue;
        for (int i = 0; i < G[v].size(); ++i) {
          edge& e = G[v][i];
          if (e.capacity > 0 && dis[e.to] > dis[v] + e.cost + H[v] - H[e.to]) {
            dis[e.to] = dis[v] + e.cost + H[v] - H[e.to];
            PreV[e.to] = v;
            PreE[e.to] = i;
            q.push(pair<int, int>(dis[e.to], e.to));
          }
        }
      }
      if (dis[t] == inf) break;
      for (int i = 0; i <= V; ++i) H[i] += dis[i];
      int d = f;
      for (int v = t; v != s; v = PreV[v])
        d = min(d, G[PreV[v]][PreE[v]].capacity);
      f -= d;
      flow += d;
      res += d * H[t];
      for (int v = t; v != s; v = PreV[v]) {
        edge& e = G[PreV[v]][PreE[v]];
        e.capacity -= d;
        G[v][e.rev].capacity += d;
      }
    }
    return res;
  }
} mcmf;
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  mcmf.Init(2 * n + 2);
  int sr = 0, tar = 2 * n + 1;
  for (int i = 1; i <= n; i++) mcmf.add(0, 2 * i - 1, 1, -1);
  for (int i = 1; i <= n; i++) mcmf.add(2 * i, tar, 1, 0);
  for (int i = 1; i <= n; i++) mcmf.add(2 * i - 1, 2 * i, 1, 0);
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if ((a[i] % 7 == a[j] % 7) or (abs(a[i] - a[j]) == 1))
        mcmf.add(2 * i, 2 * j - 1, 1, -1);
    }
  }
  int ans = mcmf.minCost(sr, tar, 4);
  cout << -ans << endl;
  return 0;
}
