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>
const long long oo = 0x3f3f3f3f;
const int _cnt = 1000 * 1000;
const int _p = 1000 * 1000 * 1000 + 7;
long long o(long long x) { return x % _p; }
int gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) { return a / gcd(a, b) * b; }
using namespace std;
void file_put() {
  freopen("filename.in", "r", stdin);
  freopen("filename.out", "w", stdout);
}
int dp[5005][5005], n, w[100005], p[15], a[100005], ans = 0;
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 0; i <= n; i++) {
    memset(p, 0, sizeof(p));
    memset(w, 0, sizeof(w));
    for (int j = 0; j <= n; j++) {
      if (j <= i)
        dp[i][j] = dp[j][i];
      else {
        dp[i][j] =
            ((dp[i][j]) > (w[a[j] - 1] + 1) ? (dp[i][j]) : (w[a[j] - 1] + 1));
        dp[i][j] =
            ((dp[i][j]) > (w[a[j] + 1] + 1) ? (dp[i][j]) : (w[a[j] + 1] + 1));
        dp[i][j] =
            ((dp[i][j]) > (p[a[j] % 7] + 1) ? (dp[i][j]) : (p[a[j] % 7] + 1));
        dp[i][j] = ((dp[i][j]) > (dp[i][0] + 1) ? (dp[i][j]) : (dp[i][0] + 1));
      }
      w[a[j]] = ((w[a[j]]) > (dp[i][j]) ? (w[a[j]]) : (dp[i][j]));
      p[a[j] % 7] = ((p[a[j] % 7]) > (dp[i][j]) ? (p[a[j] % 7]) : (dp[i][j]));
      ans = ((ans) > (dp[i][j]) ? (ans) : (dp[i][j]));
    }
  }
  printf("%d\n", ans);
  return 0;
}
