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;
struct edge {
  int to, next, from, f, c;
} e[4005 * 14];
int head[8010], dis[8010], vis[8010], q[8010], a[4005];
int from[8010], S, SS, T, tot = 1, ans;
void addE(int x, int y, int f, int c) {
  e[++tot] = (edge){y, head[x], x, f, c};
  head[x] = tot;
}
void add(int x, int y, int f, int c) {
  addE(x, y, f, c);
  addE(y, x, 0, -c);
}
bool bfs() {
  memset(dis, 63, sizeof(dis));
  memset(vis, 0, sizeof(vis));
  int h = 0, t = 1;
  q[1] = S;
  dis[S] = 0;
  while (h != t) {
    if ((++h) == 8010) h = 0;
    int x = q[h];
    vis[x] = 0;
    for (int i = head[x]; i; i = e[i].next)
      if (dis[e[i].to] > dis[x] + e[i].c && e[i].f) {
        dis[e[i].to] = dis[x] + e[i].c;
        from[e[i].to] = i;
        if (!vis[e[i].to]) {
          vis[e[i].to] = 1;
          if ((++t) == 8010) t = 0;
          q[t] = e[i].to;
        }
      }
  }
  return dis[T] < 1e8;
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  S = 0;
  SS = n * 2 + 1;
  T = n * 2 + 2;
  add(S, SS, 4, 0);
  for (int i = 1; i <= n; i++) {
    add(SS, i, 1, 0);
    add(n + i, T, 1, 0);
    add(i, n + i, 1, -1);
    add(i, n + i, 1e9, 0);
  }
  for (int i = 2; i <= n; i++) {
    for (int j = i - 1; j; j--)
      if (a[j] % 7 == a[i] % 7) {
        add(n + j, i, 1, 0);
        break;
      }
    for (int j = i - 1; j; j--)
      if (a[j] == a[i] + 1) {
        add(n + j, i, 1, 0);
        break;
      }
    for (int j = i - 1; j; j--)
      if (a[j] == a[i] - 1) {
        add(n + j, i, 1, 0);
        break;
      }
  }
  while (bfs()) {
    ans -= dis[T];
    for (int i = from[T]; i; i = from[e[i].from]) e[i].f--, e[i ^ 1].f++;
  }
  printf("%d", ans);
}
