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 maxn = 12005;
const int maxm = 100005;
const int inf = 0x3f3f3f3f;
struct MinCostMaxFlow {
  struct edge {
    int v, cap, w, rev;
    edge() {}
    edge(int v, int cap, int w, int rev) : v(v), cap(cap), w(w), rev(rev) {}
  };
  int V, s, t;
  vector<edge> G[maxn];
  inline void addedge(int u, int v, int cap, int w) {
    G[u].push_back(edge(v, cap, w, G[v].size()));
    G[v].push_back(edge(u, 0, -w, G[u].size() - 1));
  }
  int d[maxn], id[maxn], mom[maxn];
  bool inqu[maxn];
  inline int solve() {
    int mxf = 0, mnc = 0;
    while (1) {
      fill(d + 1, d + 1 + V, inf);
      fill(inqu + 1, inqu + 1 + V, 0);
      fill(mom + 1, mom + 1 + V, -1);
      mom[s] = s;
      d[s] = 0;
      queue<int> Q;
      Q.push(s);
      inqu[s] = true;
      while (!Q.empty()) {
        int u = Q.front();
        Q.pop();
        inqu[u] = false;
        for (int i = 0; i < (int)G[u].size(); ++i) {
          edge &e = G[u][i];
          int v = e.v;
          if (e.cap > 0 && d[v] > d[u] + e.w) {
            d[v] = d[u] + e.w;
            mom[v] = u;
            id[v] = i;
            if (!inqu[v]) Q.push(v), inqu[v] = true;
          }
        }
      }
      if (mom[t] == -1) break;
      int df = inf;
      for (int u = t; u != s; u = mom[u]) df = min(df, G[mom[u]][id[u]].cap);
      for (int u = t; u != s; u = mom[u]) {
        edge &e = G[mom[u]][id[u]];
        e.cap -= df;
        G[e.v][e.rev].cap += df;
      }
      mxf += df;
      mnc += df * d[t];
    }
    return mnc;
  }
} flow;
int n, a[maxn], h[7], c[maxm];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  flow.s = n + n + n + n + 2, flow.t = n + n + n + n + 3,
  flow.V = n + n + n + n + 3;
  for (int i = 0; i < 7; ++i) h[i] = flow.t;
  for (int i = 0; i < maxm; ++i) c[i] = flow.t;
  for (int i = 1; i <= n; ++i) {
    flow.addedge(flow.s, i + i, 1, 0);
    flow.addedge(i + i, i + i + 1, 1, -1);
    flow.addedge(i + i + 1, flow.t, 1, 0);
  }
  for (int i = n; i >= 1; --i) {
    int to = h[a[i] % 7];
    flow.addedge(n + n + i + i, to, inf, 0);
    flow.addedge(n + n + i + i, i + i, 1, 0);
    flow.addedge(i + i + 1, to, 1, 0);
    h[a[i] % 7] = n + n + i + i;
    to = c[a[i]];
    flow.addedge(n + n + i + i + 1, to, inf, 0);
    flow.addedge(n + n + i + i + 1, i + i, 1, 0);
    flow.addedge(i + i + 1, c[a[i] + 1], 1, 0);
    flow.addedge(i + i + 1, c[a[i] - 1], 1, 0);
    c[a[i]] = n + n + i + i + 1;
  }
  flow.s = ++flow.V;
  flow.addedge(flow.s, n + n + n + n + 2, 4, 0);
  return 0 * printf("%d", -flow.solve());
}