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>
const int maxn = 5e3 + 3, maxa = 1e5 + 3;
int n, a[maxn], dp[maxn][maxn], bkm[7], bka[maxa];
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", a + i);
  }
  int ans = 0;
  for (int i = 0; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) bka[a[j]] = 0;
    memset(bkm, 0, sizeof(bkm));
    for (int j = 1; j < i; ++j) {
      bka[a[j]] = std::max(bka[a[j]], dp[j][i]);
      bkm[a[j] % 7] = std::max(bkm[a[j] % 7], dp[j][i]);
    }
    for (int j = i + 1; j <= n; ++j) {
      dp[i][j] = std::max(std::max(dp[i][j], dp[0][i] + 1),
                          std::max(bkm[a[j] % 7] + 1,
                                   std::max(bka[a[j] + 1], bka[a[j] - 1]) + 1));
      bka[a[j]] = std::max(bka[a[j]], dp[i][j]);
      bkm[a[j] % 7] = std::max(bkm[a[j] % 7], dp[i][j]);
      ans = std::max(ans, dp[i][j]);
    }
  }
  printf("%d\n", ans);
  return 0;
}