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;
namespace TYC {
const int N = 10010;
const int inf = 0x3f3f3f3f;
int n, s, t, ss, arr[N], Head[N], pre[N], dis[N], vis[N];
struct edge {
  int to, next, w, c;
} E[N * 20];
void add(int u, int v, int w, int c) {
  static int cnt = 1;
  E[++cnt] = (edge){v, Head[u], w, c};
  Head[u] = cnt;
  E[++cnt] = (edge){u, Head[v], 0, -c};
  Head[v] = cnt;
}
bool spfa() {
  queue<int> q;
  memset(pre, 0, sizeof(pre));
  for (int i = 1; i <= t; i++) dis[i] = -inf;
  q.push(s);
  dis[s] = 0;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    vis[u] = 0;
    for (int i = Head[u]; i; i = E[i].next) {
      int v = E[i].to;
      if (E[i].w && dis[v] < dis[u] + E[i].c) {
        dis[v] = dis[u] + E[i].c;
        pre[v] = i;
        if (!vis[v]) q.push(v), vis[v] = 1;
      }
    }
  }
  return dis[t] != -inf;
}
int mcmf() {
  int ans = 0;
  while (spfa()) {
    for (int i = pre[t]; i; i = pre[E[i ^ 1].to]) E[i].w--, E[i ^ 1].w++;
    ans += dis[t];
  }
  return ans;
}
void work() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &arr[i]);
  s = n + n + 1, ss = s + 1, t = ss + 1;
  add(s, ss, 4, 0);
  for (int i = 1; i <= n; i++) {
    add(ss, i, 1, 0);
    add(i, i + n, 1, 1);
    add(i, i + n, inf, 0);
    for (int j = i + 1; j <= n; j++)
      if (arr[i] - arr[j] == 1) {
        add(i + n, j, inf, 0);
        break;
      }
    for (int j = i + 1; j <= n; j++)
      if (arr[j] - arr[i] == 1) {
        add(i + n, j, inf, 0);
        break;
      }
    for (int j = i + 1; j <= n; j++)
      if (arr[j] % 7 == arr[i] % 7) {
        add(i + n, j, inf, 0);
        break;
      }
    add(i + n, t, 1, 0);
  }
  printf("%d\n", mcmf());
}
}  // namespace TYC
int main() {
  TYC::work();
  return 0;
}
