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;
const int maxn = 5007;
inline int read() {
  int x = 0, f = 1;
  char C = getchar();
  while (!isdigit(C)) {
    if (C == '-') f = -1;
    C = getchar();
  }
  while (isdigit(C)) x = (x << 1) + (x << 3) + C - '0', C = getchar();
  return x * f;
}
int n, a[maxn], ans, f[maxn][maxn];
int mod[10], z[100007], Max;
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read(), Max = max(Max, a[i]);
  Max++;
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= Max; j++) z[j] = 0;
    for (int j = 0; j < 7; j++) mod[j] = 0;
    for (int j = 1; j <= i; j++) {
      z[a[j]] = max(z[a[j]], f[j][i]);
      mod[a[j] % 7] = max(mod[a[j] % 7], f[j][i]);
    }
    for (int j = i + 1; j <= n; j++) {
      f[i][j] = max(f[i][j], f[0][i] + 1);
      f[i][j] = max(f[i][j], mod[a[j] % 7] + 1);
      f[i][j] = max(f[i][j], max(z[a[j] + 1], z[a[j] - 1]) + 1);
      f[j][i] = f[i][j];
      mod[a[j] % 7] = max(mod[a[j] % 7], f[i][j]);
      z[a[j]] = max(z[a[j]], f[i][j]);
      ans = max(ans, f[i][j]);
    }
  }
  printf("%d\n", ans);
  return 0;
}
