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;
template <class T>
inline void read(T &x) {
  x = 0;
  register char c = getchar();
  bool f = 0;
  while (c != '-' && (c < '0' || c > '9')) c = getchar();
  if (c == '-') c = getchar(), f = 1;
  while (c >= '0' && c <= '9') x = x * 10 + c - 48, c = getchar();
  x = f ? -x : x;
  return;
}
template <class T>
inline void write(T x) {
  if (!x) {
    putchar(48);
    return;
  }
  if (x < 0) x = -x, putchar('-');
  int len = -1, z[20];
  while (x > 0) z[++len] = x % 10, x /= 10;
  for (register int i = len; i >= 0; i--) putchar(z[i] + 48);
  return;
}
const int N = 5010, M = 100010;
int f[N][N], a[N];
int pre[M], mod[10];
int main() {
  int n, ans = 0;
  read(n);
  for (int i = 1; i <= n; i++) read(a[i]);
  for (int i = 0; i <= n; i++) {
    memset(mod, 0, sizeof(mod));
    memset(pre, 0, sizeof(pre));
    for (int j = 1; j < i; j++) {
      pre[a[j]] = max(pre[a[j]], f[i][j]);
      mod[a[j] % 7] = max(mod[a[j] % 7], f[i][j]);
    }
    for (int j = i + 1; j <= n; j++) {
      f[i][j] = max(pre[a[j] - 1], pre[a[j] + 1]) + 1;
      f[i][j] = max(f[i][j], f[i][0] + 1);
      f[i][j] = max(f[i][j], mod[a[j] % 7] + 1);
      f[j][i] = f[i][j];
      pre[a[j]] = max(pre[a[j]], f[i][j]);
      mod[a[j] % 7] = max(mod[a[j] % 7], f[i][j]);
      ans = max(ans, f[i][j]);
    }
  }
  printf("%d\n", ans);
  return 0;
}
