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 INF = 0x3f3f3f3f;
const int MAXN = 3010;
int n, a[MAXN];
void input() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
}
namespace MCMF {
const int MAXV = 100010;
const int MAXE = 1000010;
struct Edge {
  int u, v, c, f, cost;
  Edge() {}
  Edge(int u, int v, int c, int f, int cost)
      : u(u), v(v), c(c), f(f), cost(cost) {}
};
int n, m, s, t;
int head[MAXV], nxt[MAXE << 1];
Edge edge[MAXE << 1];
void init(int n2) {
  n = n2, m = 0;
  for (int i = 0; i < n; ++i) head[i] = -1;
}
void adde(int u, int v, int c, int cost) {
  cost = -cost;
  edge[m] = Edge(u, v, c, 0, cost);
  nxt[m] = head[u], head[u] = m++;
  edge[m] = Edge(v, u, 0, 0, -cost);
  nxt[m] = head[v], head[v] = m++;
}
int dis[MAXV], inq[MAXV], res[MAXV], pre[MAXV];
queue<int> q;
bool spfa() {
  for (int i = 0; i < n; ++i) dis[i] = INF;
  dis[s] = 0, inq[s] = 1, res[s] = INF, q.push(s);
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    inq[u] = 0;
    for (int i = head[u]; ~i; i = nxt[i]) {
      Edge &e = edge[i];
      if (e.c > e.f && dis[e.v] > dis[u] + e.cost) {
        dis[e.v] = dis[u] + e.cost;
        pre[e.v] = i;
        res[e.v] = min(res[u], e.c - e.f);
        if (!inq[e.v]) inq[e.v] = 1, q.push(e.v);
      }
    }
  }
  return dis[t] != INF;
}
void augment() {
  int u = t, f = res[t];
  while (u != s) {
    int i = pre[u];
    edge[i].f += f;
    edge[i ^ 1].f -= f;
    u = edge[i].u;
  }
}
int solve(int s2, int t2) {
  s = s2, t = t2;
  int cost = 0;
  while (spfa()) {
    cost += dis[t];
    augment();
  }
  return -cost;
}
}  // namespace MCMF
int id(int i, int j) { return i * n + j; }
void solve() {
  int s = id(6, 0), t = id(6, 1), ss = id(6, 2);
  MCMF::init(ss + 1);
  MCMF::adde(ss, s, 4, 0);
  for (int i = 0; i < n; ++i) {
    MCMF::adde(id(0, i), id(1, i), 1, 1);
    MCMF::adde(s, id(0, i), 1, 0);
    MCMF::adde(id(1, i), t, 1, 0);
    MCMF::adde(id(2, i), id(0, i), 1, 0);
    MCMF::adde(id(4, i), id(0, i), 1, 0);
    MCMF::adde(id(1, i), id(3, i), 1, 0);
    MCMF::adde(id(1, i), id(5, i), 1, 0);
  }
  for (int i = 0; i < n; ++i) {
    for (int j = i + 1; j < n; ++j)
      if (a[i] % 7 == a[j] % 7) {
        MCMF::adde(id(2, i), id(2, j), INF, 0);
        MCMF::adde(id(3, i), id(2, j), INF, 0);
        break;
      }
    for (int j = i + 1; j < n; ++j)
      if (a[i] == a[j]) {
        MCMF::adde(id(4, i), id(4, j), INF, 0);
        break;
      }
    for (int j = i + 1; j < n; ++j)
      if (a[j] == a[i] + 1) {
        MCMF::adde(id(5, i), id(4, j), INF, 0);
        break;
      }
    for (int j = i + 1; j < n; ++j)
      if (a[j] == a[i] - 1) {
        MCMF::adde(id(5, i), id(4, j), INF, 0);
        break;
      }
  }
  printf("%d\n", MCMF::solve(ss, t));
}
int main() {
  input(), solve();
  return 0;
}