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;
const int maxv = 1e4 + 5, maxn = 1e5 + 5;
const int inf = 1e9 + 5;
struct MCMF {
  struct Edge {
    int to, rev, cap, w;
    Edge() {}
    Edge(int a, int b, int c, int d) : to(a), cap(b), w(c), rev(d) {}
  };
  int n, s, t;
  vector<int> p, id, d;
  bitset<maxn> inque;
  vector<Edge> G[maxn];
  pair<int, int> spfa() {
    p.assign(n, -1);
    d.assign(n, inf);
    id.assign(n, -1);
    d[s] = 0;
    p[s] = s;
    queue<int> que;
    que.push(s);
    inque[s] = true;
    while (que.size()) {
      int tmp = que.front();
      que.pop();
      inque[tmp] = false;
      int i = 0;
      for (auto e : G[tmp]) {
        if (e.cap > 0 && d[e.to] > d[tmp] + e.w) {
          d[e.to] = d[tmp] + e.w;
          p[e.to] = tmp;
          id[e.to] = i;
          if (!inque[e.to]) que.push(e.to), inque[e.to] = true;
        }
        ++i;
      }
    }
    if (d[t] == inf) return make_pair(-1, -1);
    int a = inf;
    for (int i = t; i != s; i = p[i]) {
      a = min(a, G[p[i]][id[i]].cap);
    }
    for (int i = t; i != s; i = p[i]) {
      Edge &e = G[p[i]][id[i]];
      e.cap -= a;
      G[e.to][e.rev].cap += a;
    }
    return make_pair(a, d[t]);
  }
  MCMF(int _n, int _s, int _t) : n(_n), s(_s), t(_t) {
    fill(G, G + maxn, vector<Edge>());
  }
  void add_edge(int a, int b, int cap, int w) {
    G[a].push_back(Edge(b, cap, w, (int)G[b].size()));
    G[b].push_back(Edge(a, 0, -w, (int)G[a].size() - 1));
  }
  pair<int, int> maxflow() {
    int mxf = 0, mnc = 0;
    while (true) {
      pair<int, int> res = spfa();
      if (res.first == -1) break;
      mxf += res.first;
      mnc += res.first * res.second;
    }
    return make_pair(mxf, mnc);
  }
};
int a[maxn], nxt[maxn];
int z[7];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  MCMF flow(n + n + 10, n + n + 2, n + n + 1);
  for (int i = 1; i <= n; ++i)
    flow.add_edge(i, i + n, 1, -1), flow.add_edge(i, i + n, inf, 0);
  for (int i = n; i >= 1; --i) {
    if (nxt[a[i] + 1]) flow.add_edge(i + n, nxt[a[i] + 1], 1, 0);
    if (nxt[a[i] - 1]) flow.add_edge(i + n, nxt[a[i] - 1], 1, 0);
    if (z[a[i] % 7]) flow.add_edge(i + n, z[a[i] % 7], 4, 0);
    z[a[i] % 7] = i;
    nxt[a[i]] = i;
  }
  for (int i = 1; i <= n; ++i) flow.add_edge(0, i, 1, 0);
  for (int i = 1; i <= n; ++i) flow.add_edge(i + n, n + n + 1, 1, 0);
  flow.add_edge(n + n + 2, 0, 4, 0);
  int f = 0, c = 0;
  tie(f, c) = flow.maxflow();
  printf("%d\n", -c);
}
