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 n;
int a[5001];
class MCMF {
 public:
  static const int sz = 20000;
  class edge {
   public:
    using pointer = edge*;
    int v, c, w;
    pointer rev;
    edge(int v, int c, int w) : v(v), c(c), w(w) {}
  };
  vector<edge::pointer> g[sz];
  int n;
  void reset(int _n) {
    n = _n;
    for (int i = 0; i < n; i++) g[i].clear();
  }
  void add_edge(int u, int v, int c, int w) {
    g[u].push_back(new edge(v, c, w));
    g[v].push_back(new edge(u, 0, -w));
    g[u].back()->rev = g[v].back();
    g[v].back()->rev = g[u].back();
  }
  int d[sz];
  edge::pointer trace[sz];
  int source, sink;
  queue<int> q;
  bool inq[sz];
  bool find_path() {
    for (int i = 0; i < sz; i++) d[i] = 1e9;
    d[source] = 0;
    q.push(source);
    inq[source] = 1;
    while (!q.empty()) {
      auto u = q.front();
      inq[u] = 0;
      q.pop();
      for (auto i : g[u])
        if (i->c)
          if (d[i->v] > d[u] + i->w) {
            d[i->v] = d[u] + i->w;
            trace[i->v] = i;
            if (!inq[i->v]) {
              inq[i->v] = 1;
              q.push(i->v);
            }
          }
    }
    return d[sink] != 1e9;
  }
  int ansflow, anscost;
  void increase_flow() {
    int u = sink;
    int cap = 1e9;
    while (u != source) {
      cap = min(cap, trace[u]->c);
      u = trace[u]->rev->v;
    }
    ansflow += cap;
    anscost += cap * d[sink];
    u = sink;
    while (u != source) {
      trace[u]->c -= cap;
      trace[u]->rev->c += cap;
      u = trace[u]->rev->v;
    }
  }
  pair<int, int> flow(int s, int t) {
    source = s;
    sink = t;
    ansflow = anscost = 0;
    while (find_path()) increase_flow();
    return {ansflow, anscost};
  }
} F;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  F.reset(n * 2 + 3);
  for (int i = 1; i <= n; i++) F.add_edge(i + 2, i + n + 2, 1, -1);
  for (int i = 1; i <= n; i++) F.add_edge(1, i + 2, 1, 0);
  for (int i = 1; i <= n; i++) F.add_edge(i + n + 2, 2, 1, 0);
  const int flowsize = 4;
  for (int i = 1; i <= n; i++) {
    int mod = flowsize, p1 = flowsize, n1 = flowsize;
    for (int j = i + 1; j <= n; j++) {
      if ((a[i] - a[j]) % 7 == 0) {
        if (mod) {
          mod--;
          F.add_edge(i + n + 2, j + 2, 1, 0);
        }
      } else if (a[j] == a[i] + 1) {
        if (p1) {
          p1--;
          F.add_edge(i + n + 2, j + 2, 1, 0);
        }
      } else if (a[j] == a[i] - 1) {
        if (n1) {
          n1--;
          F.add_edge(i + n + 2, j + 2, 1, 0);
        }
      }
    }
  }
  F.add_edge(0, 1, flowsize, 0);
  cout << -F.flow(0, 2).second << '\n';
}
