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, flow, cost, nxt;
} e[100050];
int now, S, SS, T, maxcost, first[10050], f[10050], a[10050], path[10050],
    number[10050], q[10000050];
bool inq[10050];
void Add(int x, int y, int flow, int cost) {
  e[++now].to = y;
  e[now].flow = flow;
  e[now].cost = cost;
  e[now].nxt = first[x];
  first[x] = now;
}
bool Find() {
  int low, high, pt, nxt;
  memset(f, 128, sizeof(f));
  memset(inq, false, sizeof(inq));
  f[S] = 0;
  inq[S] = true;
  q[low = high = 0] = S;
  for (; low <= high; inq[q[low++]] = false)
    for (pt = first[q[low]]; pt != -1; pt = e[pt].nxt) {
      nxt = e[pt].to;
      if (e[pt].flow <= 0 || f[nxt] >= f[q[low]] + e[pt].cost || q[low] == T ||
          nxt == S)
        continue;
      f[nxt] = f[q[low]] + e[pt].cost;
      path[nxt] = q[low];
      number[nxt] = pt;
      if (!inq[nxt]) {
        inq[nxt] = true;
        q[++high] = nxt;
      }
    }
  return f[T] > 0;
}
void Addflow() {
  int i, t;
  for (i = T; i != S; i = path[i]) t = min(t, e[number[i]].flow);
  maxcost += f[T] * t;
  for (i = T; i != S; i = path[i]) {
    e[number[i]].flow -= t;
    e[number[i] ^ 1].flow += t;
  }
}
int main() {
  int n, i, j;
  scanf("%d", &n);
  for (i = 1; i <= n; i++) scanf("%d", &a[i]);
  S = 0;
  SS = 2 * n + 1;
  T = 2 * n + 2;
  memset(first, -1, sizeof(first));
  now = -1;
  Add(S, SS, 4, 0);
  Add(SS, S, 0, 0);
  for (i = 1; i <= n; i++) {
    Add(SS, i, 1, 0);
    Add(i, SS, 0, 0);
    Add(i, n + i, 1, 1);
    Add(n + i, i, 0, -1);
    Add(n + i, T, 1, 0);
    Add(T, n + i, 0, 0);
  }
  for (i = 1; i <= n; i++) {
    for (j = i + 1; j <= n; j++)
      if (a[i] % 7 == a[j] % 7) {
        Add(n + i, j, 1, 0);
        Add(j, n + i, 0, 0);
        Add(i, j, 1, 0);
        Add(j, i, 0, 0);
        break;
      }
    for (j = i + 1; j <= n; j++)
      if (a[i] + 1 == a[j]) {
        Add(n + i, j, 1, 0);
        Add(j, n + i, 0, 0);
        Add(i, j, 1, 0);
        Add(j, i, 0, 0);
        break;
      }
    for (j = i + 1; j <= n; j++)
      if (a[j] + 1 == a[i]) {
        Add(n + i, j, 1, 0);
        Add(j, n + i, 0, 0);
        Add(i, j, 1, 0);
        Add(j, i, 0, 0);
        break;
      }
  }
  while (Find()) Addflow();
  printf("%d\n", maxcost);
  return 0;
}
