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 nmax = 1e4 + 5, inf = 1e9;
struct edge {
  int from, to, flow, cost;
};
vector<int> adj[nmax];
vector<edge> all;
void add_edge(int from, int to, int flow, int cost) {
  edge current;
  current.from = from;
  current.to = to;
  current.flow = flow;
  current.cost = cost;
  adj[from].push_back(all.size());
  all.push_back(current);
  swap(current.from, current.to);
  current.flow = 0;
  current.cost = -cost;
  adj[to].push_back(all.size());
  all.push_back(current);
}
int source, sink;
int n, k;
int parent[nmax];
long long cost[nmax];
int type[nmax];
deque<int> active;
void spfa() {
  for (int i = source; i <= sink; i++) {
    parent[i] = -1;
    cost[i] = 1e18;
    type[i] = 2;
  }
  parent[source] = 0;
  cost[source] = 0;
  type[source] = 1;
  active.push_back(source);
  while (active.size()) {
    int current = active.front();
    active.pop_front();
    for (auto k : adj[current])
      if (all[k].flow && cost[all[k].from] + all[k].cost < cost[all[k].to]) {
        cost[all[k].to] = cost[all[k].from] + all[k].cost;
        parent[all[k].to] = k;
        if (type[all[k].to] == 1) continue;
        if (active.size() && cost[all[k].to] < cost[active.front()])
          active.push_front(all[k].to);
        else
          active.push_back(all[k].to);
        type[all[k].to] = 1;
      }
    type[current] = 0;
  }
}
long long min_cost_max_flow(int wanted) {
  int total_flow = 0;
  long long total_cost = 0;
  while (total_flow < wanted) {
    spfa();
    if (parent[sink] == -1) break;
    if (cost[sink] >= 0) break;
    int node = sink;
    long long cost_now = 0;
    while (node != source) {
      int used_edge = parent[node];
      all[used_edge].flow--;
      all[used_edge ^ 1].flow++;
      cost_now += all[used_edge].cost;
      node = all[used_edge].from;
    }
    total_flow++;
    total_cost = total_cost + cost_now;
  }
  return total_cost;
}
int arr[nmax];
int main() {
  scanf("%i", &n);
  for (int i = 1; i <= n; i++) scanf("%i", &arr[i]);
  source = 0;
  sink = 2 * n + 1;
  for (int i = 1; i <= n; i++) add_edge(source, 2 * i - 1, 1, 0);
  for (int j = 1; j <= n; j++) add_edge(2 * j, sink, 1, 0);
  for (int i = 1; i <= n; i++) add_edge(2 * i - 1, 2 * i, 1, -1);
  for (int j = 1; j <= n; j++)
    for (int i = j + 1; i <= n; i++)
      if (arr[i] % 7 == arr[j] % 7 || abs(arr[i] - arr[j]) == 1)
        add_edge(2 * j, 2 * i - 1, 1, 0);
  printf("%lld\n", -min_cost_max_flow(4));
  return 0;
}
