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;
int n, num[3010], s;
namespace MCMF {
const int N = 15000, M = 2000000;
int head[N], cnt, dis[N], fr[N], id[N], S, T, cost;
struct node {
  int to, next, val, cost;
} edge[M];
void ae(int u, int v, int w, int c) {
  edge[cnt].cost = c, edge[cnt].next = head[u], edge[cnt].to = v,
  edge[cnt].val = w, head[u] = cnt++;
  edge[cnt].cost = -c, edge[cnt].next = head[v], edge[cnt].to = u,
  edge[cnt].val = 0, head[v] = cnt++;
}
queue<int> q;
bool in[N];
bool SPFA() {
  memset(dis, -1, sizeof(dis)), dis[S] = 0, q.push(S), in[S] = true;
  while (!q.empty()) {
    int x = q.front();
    q.pop(), in[x] = false;
    for (int i = head[x]; i != -1; i = edge[i].next) {
      if (!edge[i].val) continue;
      if (dis[edge[i].to] < dis[x] + edge[i].cost) {
        dis[edge[i].to] = dis[x] + edge[i].cost, fr[edge[i].to] = x,
        id[edge[i].to] = i;
        if (!in[edge[i].to]) in[edge[i].to] = true, q.push(edge[i].to);
      }
    }
  }
  if (dis[T] == -1) return false;
  int x = T, mn = 0x3f3f3f3f;
  while (x != S) mn = min(mn, edge[id[x]].val), x = fr[x];
  cost += dis[T] * mn, x = T;
  while (x != S) edge[id[x]].val -= mn, edge[id[x] ^ 1].val += mn, x = fr[x];
  return true;
}
}  // namespace MCMF
using namespace MCMF;
int main() {
  scanf("%d", &n), memset(head, -1, sizeof(head)),
      S = 4 * n + 1, T = 4 * n + 2, s = 4 * n + 3, ae(S, s, 4, 0);
  for (int i = 1; i <= n; i++) scanf("%d", &num[i]);
  for (int i = 1; i <= n; i++) {
    ae(s, i + 2 * n, 0x3f3f3f3f, 0);
    ae(i + 0 * n, i + 2 * n, 0x3f3f3f3f, 0);
    ae(i + 1 * n, i + 2 * n, 0x3f3f3f3f, 0);
    ae(i + 2 * n, i + 3 * n, 1, 1);
    ae(i + 3 * n, T, 0x3f3f3f3f, 0);
    for (int j = i + 1; j <= n; j++)
      if (num[i] - num[j] == 1) {
        ae(i + 3 * n, j + 1 * n, 0x3f3f3f3f, 0);
        break;
      }
    for (int j = i + 1; j <= n; j++)
      if (num[j] - num[i] == 1) {
        ae(i + 3 * n, j + 1 * n, 0x3f3f3f3f, 0);
        break;
      }
    for (int j = i + 1; j <= n; j++)
      if (num[i] % 7 == num[j] % 7) {
        ae(i + 3 * n, j + 0 * n, 0x3f3f3f3f, 0);
        break;
      }
    for (int j = i + 1; j <= n; j++)
      if (num[i] % 7 == num[j] % 7) {
        ae(i + 0 * n, j + 0 * n, 0x3f3f3f3f, 0);
        break;
      }
    for (int j = i + 1; j <= n; j++)
      if (num[i] == num[j]) {
        ae(i + 1 * n, j + 1 * n, 0x3f3f3f3f, 0);
        break;
      }
  }
  while (SPFA())
    ;
  printf("%d\n", cost);
  return 0;
}
