Problem: 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).

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int x = 0, f = 1;
  char ch = getchar();
  while (!(ch >= '0' && ch <= '9')) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    x = x * 10 + (ch - '0');
    ch = getchar();
  }
  return x * f;
}
const int N = 5005;
const int M = 100005;
const int T = 7;
int n, a[N];
int ans = 0;
int dp[N][N];
int mx1[M], mx2[T];
int main() {
  n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  dp[0][0] = 0;
  for (int j = 0; j <= n; j++) {
    for (int i = 1; i <= j - 1; i++) {
      mx1[a[i]] = max(mx1[a[i]], dp[j][i]);
      mx2[a[i] % 7] = max(mx2[a[i] % 7], dp[j][i]);
    }
    for (int i = j + 1; i <= n; i++) {
      dp[i][j] =
          max(max(mx1[a[i] - 1], mx1[a[i] + 1]), max(dp[j][0], mx2[a[i] % 7])) +
          1;
      mx1[a[i]] = max(mx1[a[i]], dp[i][j]);
      mx2[a[i] % 7] = max(mx2[a[i] % 7], dp[i][j]);
      ans = max(ans, dp[i][j]);
    }
    for (int i = 1; i <= n; i++) mx1[a[i]] = 0;
    for (int i = 0; i <= 6; i++) mx2[i] = 0;
  }
  printf("%d\n", ans);
  return 0;
}