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 = 6005;
const int Inf = 1000000000;
struct edge {
  int a, b, cap, cost;
  edge(int a = 0, int b = 0, int cap = 0, int cost = 0)
      : a(a), b(b), cap(cap), cost(cost) {}
};
int n;
int a[Maxn];
int N;
vector<edge> E;
vector<int> neigh[Maxn];
int val[Maxn], par[Maxn];
void addEdge(int a, int b, int cap, int cost) {
  neigh[a].push_back(E.size());
  E.push_back(edge(a, b, cap, cost));
  neigh[b].push_back(E.size());
  E.push_back(edge(b, a, 0, -cost));
}
int Flow() {
  fill(val, val + Maxn, -Inf);
  val[0] = 0;
  priority_queue<pair<int, int> > Q;
  Q.push(pair<int, int>(val[0], -0));
  while (!Q.empty()) {
    int v = -Q.top().second;
    int d = Q.top().first;
    Q.pop();
    if (val[v] != d) continue;
    for (int i = 0; i < neigh[v].size(); i++) {
      int e = neigh[v][i];
      if (E[e].cap > 0 && d + E[e].cost > val[E[e].b]) {
        val[E[e].b] = d + E[e].cost;
        par[E[e].b] = e;
        Q.push(pair<int, int>(val[E[e].b], -E[e].b));
      }
    }
  }
  int v = N - 1;
  int res = val[v];
  while (v) {
    int e = par[v];
    E[e].cap--;
    E[e ^ 1].cap++;
    v = E[e].a;
  }
  return res;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  N = 2 * n + 2;
  for (int i = 1; i <= n; i++) {
    addEdge(0, 2 * i - 1, 1, 0);
    addEdge(2 * i, N - 1, 1, 0);
    addEdge(2 * i - 1, 2 * i, 1, 1);
  }
  for (int i = 1; i <= n; i++)
    for (int j = i + 1; j <= n; j++)
      if (a[i] % 7 == a[j] % 7 || abs(a[i] - a[j]) == 1)
        addEdge(2 * i, 2 * j - 1, 1, 0);
  int res = 0;
  for (int i = 0; i < 4; i++) res += Flow();
  printf("%d\n", res);
  return 0;
}