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;
class Edge {
 public:
  int from, to, flow, cost;
  Edge(int _from = 0, int _to = 0, int _flow = 0, int _cost = 0)
      : from(_from), to(_to), flow(_flow), cost(_cost) {}
};
vector<Edge> edges;
vector<vector<int>> adia;
vector<int> dist, potential;
vector<int> father;
int S, D, n;
bool djk() {
  potential = dist;
  dist = vector<int>(n, 1e9);
  father = vector<int>(n, 1e9);
  priority_queue<pair<int, int>> q;
  q.push({0, S});
  dist[S] = 0;
  while (!q.empty()) {
    int t(-q.top().first), x(q.top().second);
    q.pop();
    if (dist[x] != t) continue;
    for (auto i : adia[x]) {
      if (!edges[i].flow) continue;
      int c(t + potential[x] - potential[edges[i].to] + edges[i].cost);
      if (dist[edges[i].to] > c) {
        dist[edges[i].to] = c;
        father[edges[i].to] = i;
        q.push({-c, edges[i].to});
      }
    }
  }
  for (int i(0); i < dist.size(); i++) dist[i] += potential[i];
  return father[D] != 1e9;
}
int recursive_min_dist(int nod) {
  if (dist[nod] != 1e9) return dist[nod];
  dist[nod]--;
  for (auto i : adia[nod])
    if (edges[i ^ 1].flow)
      dist[nod] =
          min(dist[nod], recursive_min_dist(edges[i].to) + edges[i ^ 1].cost);
  return dist[nod];
}
int min_cost_max_flow() {
  int cost(0);
  dist = vector<int>(n, 1e9);
  dist[S] = 0;
  for (int i(0); i < n; i++) recursive_min_dist(i);
  while (djk()) {
    int dmin(1e9), sumcost(0);
    for (int i(D); i != S; i = edges[father[i]].from)
      dmin = min(dmin, edges[father[i]].flow), sumcost += edges[father[i]].cost;
    cost += dmin * sumcost;
    for (int i(D); i != S; i = edges[father[i]].from)
      edges[father[i]].flow -= dmin, edges[father[i] ^ 1].flow += dmin;
  }
  return cost;
}
void add_edge(Edge x) {
  adia[x.from].push_back(edges.size());
  edges.push_back(x);
  adia[x.to].push_back(edges.size());
  edges.push_back({x.to, x.from, 0, -x.cost});
}
void Graph(int _n = 0, int _s = 0, int _d = 0) {
  S = _s, D = _d;
  n = _n;
  adia.resize(n);
}
int main() {
  int n;
  vector<int> v;
  cin >> n;
  v.resize(n + 1);
  for (int i(1); i <= n; i++) cin >> v[i];
  Graph(2 * n + 3, 0, 2 * n + 2);
  add_edge({2 * n + 1, 2 * n + 2, 2, 0});
  for (int i(1); i <= n; i++) {
    add_edge({0, i, 1, 0});
    add_edge({i, i + n, 1, -1});
    add_edge({i + n, 2 * n + 1, 1, 0});
  }
  for (int i(1); i < n; i++) {
    int cnt(0);
    for (int j(i + 1); j <= n; j++) {
      if (cnt <= 100 && v[i] % 7 == v[j] % 7 || v[i] == v[j] + 1 ||
          v[i] == v[j] - 1)
        add_edge({i + n, j, 1, 0}), cnt++;
    }
  }
  cout << -min_cost_max_flow();
  return 0;
}