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 mod = 1e9 + 7;
const int inf = 1e9;
struct edge {
  int to, cap, cost, rev;
};
const int MAX_V = 6005, INF = 1e9;
int V;
vector<edge> G[MAX_V];
int h[MAX_V];
int dist[MAX_V];
int prevv[MAX_V], preve[MAX_V];
void add_edge(int from, int to, int cap, int cost) {
  edge e1 = {to, cap, cost, (int)G[to].size()};
  edge e2 = {from, 0, -cost, (int)G[from].size()};
  G[from].push_back(e1);
  G[to].push_back(e2);
}
int min_cost_flow(int s, int t, int f) {
  int res = 0;
  fill(h, h + V, 0);
  while (f > 0) {
    priority_queue<pair<int, int>, vector<pair<int, int> >,
                   greater<pair<int, int> > >
        que;
    fill(dist, dist + V, INF);
    dist[s] = 0;
    que.push(pair<int, int>(0, s));
    while (!que.empty()) {
      pair<int, int> p = que.top();
      que.pop();
      int v = p.second;
      if (dist[v] < p.first) continue;
      for (int i = 0; i < G[v].size(); i++) {
        edge &e = G[v][i];
        if (e.cap > 0 && dist[e.to] > dist[v] + e.cost + h[v] - h[e.to]) {
          dist[e.to] = dist[v] + e.cost + h[v] - h[e.to];
          prevv[e.to] = v;
          preve[e.to] = i;
          que.push(pair<int, int>(dist[e.to], e.to));
        }
      }
    }
    if (dist[t] == INF) return res;
    for (int v = 0; v < V; v++) h[v] += dist[v];
    int d = f;
    for (int v = t; v != s; v = prevv[v]) {
      d = min(d, G[prevv[v]][preve[v]].cap);
    }
    f -= d;
    res += d * h[t];
    for (int v = t; v != s; v = prevv[v]) {
      edge &e = G[prevv[v]][preve[v]];
      e.cap -= d;
      G[v][e.rev].cap += d;
    }
  }
  return res;
}
void reset() {
  for (int i = 0; i < V; i++) G[i].clear();
}
int a[3005];
signed main() {
  int n;
  scanf("%d", &n);
  V = 2 * n + 3;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  add_edge(0, 2 * n + 1, 4, 0);
  for (int i = 1; i <= n; i++) {
    add_edge(i, n + i, 1, -1);
    add_edge(2 * n + 1, i, 1, 0);
    add_edge(n + i, 2 * n + 2, 1, 0);
    for (int j = i + 1; j <= n; j++) {
      if ((a[i] % 7 == a[j] % 7) || abs(a[i] - a[j]) == 1) {
        add_edge(i + n, j, 1, 0);
      }
    }
  }
  printf("%d", -min_cost_flow(0, 2 * n + 2, inf));
  return 0;
}