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>
int n, a[5010], b[5010], *cnt, r[5010], pre[5010], F[100010], G[5010],
    f[5010][5010], g[7][5010];
void cmax(int& a, int b) { b > a ? a = b : 1; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i), b[i] = a[i];
  std::sort(b + 1, b + n + 1);
  cnt = std::unique(b + 1, b + n + 1);
  for (int i = 1; i <= n; i++) r[i] = std::lower_bound(b + 1, cnt, a[i]) - b;
  int s = 0;
  for (int i = 1; i <= n; i++) {
    cmax(pre[i], F[a[i] - 1]);
    cmax(pre[i], F[a[i] + 1]);
    cmax(pre[i], G[a[i] % 7]);
    pre[i]++;
    cmax(F[a[i]], pre[i]);
    cmax(G[a[i] % 7], pre[i]);
  }
  for (int i = n; i; i--) {
    for (int j = i; --j;) {
      int t = 0;
      if (b[r[i] - 1] == b[r[i]] - 1) cmax(t, f[r[i] - 1][j]);
      if (b[r[i] + 1] == b[r[i]] + 1) cmax(t, f[r[i] + 1][j]);
      if (b[r[j] - 1] == b[r[j]] - 1) cmax(t, f[r[j] - 1][i]);
      if (b[r[j] + 1] == b[r[j]] + 1) cmax(t, f[r[j] + 1][i]);
      cmax(t, g[a[i] % 7][j]);
      cmax(t, g[a[j] % 7][i]);
      cmax(f[r[i]][j], ++t);
      cmax(g[a[i] % 7][j], t);
      cmax(s, t + pre[j]);
    }
  }
  printf("%d\n", s);
}
