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 = 1e9 + 10, MN = 5e4 + 10, MV = 1e5 + 10, MS = 1e7 + 5;
int n, m, arr[MN], lst[MV];
int prv[MN], head[MN], from[MN], to[MN], cap[MN], w[MN];
int par[MN], dist[MN];
bool isinq[MN];
int L, R, q[MS];
void Add(int x) {
  isinq[x] = true;
  q[R++] = x;
  if (R == MS) R = 0;
}
int Pop() {
  int ret = q[L++];
  isinq[ret] = false;
  if (L == MS) L = 0;
  return ret;
}
void add_edge(int u, int v, int Cap, int W) {
  from[m] = u, to[m] = v, cap[m] = Cap, w[m] = W;
  prv[m] = head[u], head[u] = m++;
  from[m] = v, to[m] = u, cap[m] = 0, w[m] = -W;
  prv[m] = head[v], head[v] = m++;
}
int min_cost(int st, int en) {
  fill(dist, dist + n, INF);
  dist[st] = 0;
  Add(st);
  while (L != R) {
    int v = Pop();
    for (int e = head[v]; ~e; e = prv[e]) {
      int u = to[e];
      if (cap[e] && dist[v] + w[e] < dist[u]) {
        dist[u] = dist[v] + w[e];
        par[u] = e;
        if (!isinq[u]) Add(u);
      }
    }
  }
  if (dist[en] >= 0) return 0;
  int cur = en;
  while (cur ^ st) {
    int e = par[cur];
    cap[e] -= 1;
    cap[e ^ 1] += 1;
    cur = from[e];
  }
  return dist[en] + min_cost(st, en);
}
int main() {
  ios_base ::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> arr[i];
  memset(head, -1, sizeof head);
  memset(prv, -1, sizeof prv);
  memset(lst, -1, sizeof lst);
  for (int i = n - 1; ~i; --i) {
    if (~lst[arr[i] + 1]) {
      int x = lst[arr[i] + 1];
      add_edge(2 * i + 1, 2 * x, 1, 0);
    }
    if (~lst[arr[i] - 1]) {
      int y = lst[arr[i] - 1];
      add_edge(2 * i + 1, 2 * y, 1, 0);
    }
    add_edge(2 * i + 1, 2 * n, 1, 0);
    add_edge(2 * n + 1, 2 * i, 1, 0);
    add_edge(2 * i, 2 * i + 1, 1, -1);
    lst[arr[i]] = i;
  }
  add_edge(2 * n + 2, 2 * n + 1, 4, 0);
  int sz = 2 * n + 3;
  memset(lst, -1, sizeof lst);
  for (int i = n - 1; ~i; --i) {
    int x = arr[i] % 7;
    add_edge(sz, 2 * i, 1, 0);
    if (~lst[x]) {
      add_edge(2 * i + 1, lst[x], 1, 0);
      add_edge(sz, lst[x], 1, 0);
    }
    lst[x] = sz++;
  }
  int st = 2 * n + 2, en = 2 * n;
  n = sz;
  cout << -min_cost(st, en) << '\n';
  return 0;
}