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;
using T = int;
struct Edge {
  int from, to;
  T cost, flow, cap;
  Edge(int _from = 0, int _to = 0, T _cost = 0, T _cap = 0)
      : from(_from), to(_to), cost(_cost), flow(0), cap(_cap) {}
};
const int N = 7010;
const int INF = 0x3f3f3f3f;
int n;
vector<Edge> edges;
vector<int> g[N];
int par[N];
int dist[N];
int pi[N];
void addEdge(int from, int to, T cost, T cap) {
  g[from].push_back(edges.size());
  edges.emplace_back(from, to, cost, cap);
  g[to].push_back(edges.size());
  edges.emplace_back(to, from, -cost, 0);
}
bool dijkstra(int s, int t) {
  for (int i = 0; i <= t; ++i) {
    dist[i] = INF;
    par[i] = -1;
  }
  priority_queue<pair<T, int>, vector<pair<T, int>>, greater<pair<T, int>>> pq;
  dist[s] = 0;
  pq.push({0, s});
  while (!pq.empty()) {
    auto on = pq.top();
    pq.pop();
    int chosen = on.second;
    int d = on.first;
    if (d > dist[chosen]) continue;
    for (int edge_id : g[chosen]) {
      int to = edges[edge_id].to;
      int cost = edges[edge_id].cost;
      T flowing = edges[edge_id].flow;
      T cap = edges[edge_id].cap;
      if (cap <= flowing) continue;
      if (dist[to] > dist[chosen] + pi[chosen] - pi[to] + cost) {
        dist[to] = dist[chosen] + pi[chosen] - pi[to] + cost;
        par[to] = edge_id;
        pq.push({dist[to], to});
      }
    }
  }
  return dist[t] != INF;
}
int mcmf(int s, int t) {
  memset(pi, 0, sizeof pi);
  int cost = 0, max_flow = 0;
  while (dijkstra(s, t)) {
    for (int i = 0; i <= t; ++i) {
      pi[i] += dist[i];
    }
    int cur = t, flow = (1 << 30);
    for (int cur = t; par[cur] != -1; cur = edges[par[cur]].from) {
      flow = min(flow, edges[par[cur]].cap - edges[par[cur]].flow);
    }
    for (int cur = t; par[cur] != -1; cur = edges[par[cur]].from) {
      cost += edges[par[cur]].cost * flow;
      edges[par[cur]].flow += flow;
      edges[par[cur] ^ 1].flow -= flow;
    }
    max_flow += flow;
  }
  return cost;
}
int main() {
  scanf("%d", &n);
  vector<int> a(n);
  addEdge(0, 1, 0, 4);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < n; ++i) {
    addEdge(1, i + 2, 0, 1);
    addEdge(i + 2, n + i + 2, -1, 1);
    addEdge(n + i + 2, 2 * n + 2, 0, 1);
    for (int j = i + 1; j < n; ++j) {
      if (abs(a[i] - a[j]) == 1 or a[i] % 7 == a[j] % 7) {
        addEdge(n + i + 2, j + 2, 0, 1);
      }
    }
  }
  printf("%d\n", -1 * mcmf(0, 2 * n + 2));
  return 0;
}