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;
long long n;
long long a[5005], f[5005][5005];
long long ans, p1[100005], p2[10];
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') f = -1;
    ch = getchar();
    if (ch == -1) return 0;
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
inline long long max_(long long a, long long b, long long c, long long d) {
  return max(max(a, b), max(c, d));
}
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) a[i] = read();
  for (long long i = 0; i <= n; i++) {
    memset(p1, 0, sizeof(p1)), memset(p2, 0, sizeof(p2));
    for (long long j = 1; j <= i - 1; j++)
      p1[a[j]] = max(p1[a[j]], f[i][j]),
      p2[a[j] % 7] = max(p2[a[j] % 7], f[i][j]);
    for (long long j = i + 1; j <= n; j++) {
      f[i][j] = f[j][i] = max_(f[i][0] + 1, p1[a[j] + 1] + 1, p1[a[j] - 1] + 1,
                               p2[a[j] % 7] + 1);
      p1[a[j]] = max(p1[a[j]], f[i][j]),
      p2[a[j] % 7] = max(p2[a[j] % 7], f[i][j]), ans = max(ans, f[i][j]);
    }
  }
  cout << ans << endl;
  return 0;
}
