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 <typename T>
void get_int(T &x) {
  char t = getchar();
  bool neg = false;
  x = 0;
  for (; (t > '9' || t < '0') && t != '-'; t = getchar())
    ;
  if (t == '-') neg = true, t = getchar();
  for (; t <= '9' && t >= '0'; t = getchar()) x = x * 10 + t - '0';
  if (neg) x = -x;
}
template <typename T>
void print_int(T x) {
  if (x < 0) putchar('-'), x = -x;
  short a[20] = {}, sz = 0;
  while (x > 0) a[sz++] = x % 10, x /= 10;
  if (sz == 0) putchar('0');
  for (int i = sz - 1; i >= 0; i--) putchar('0' + a[i]);
}
const int inf = 0x3f3f3f3f;
const long long Linf = 1ll << 61;
const double pi = acos(-1.0);
int n, a[5111], dp[5111][5111], mx1[100111], mxm[8], ans;
int main() {
  get_int(n);
  for (int i = 1; i <= n; i++) get_int(a[i]);
  for (int i = 0; i <= n; i++) {
    memset(mxm, 0, sizeof(mxm));
    for (int j = 1; j <= n; j++) {
      if (j > i)
        dp[i][j] = dp[j][i] = max(max(dp[i][0], mxm[a[j] % 7]),
                                  max(mx1[a[j] - 1], mx1[a[j] + 1])) +
                              1;
      mx1[a[j]] = max(mx1[a[j]], dp[i][j]);
      mxm[a[j] % 7] = max(mxm[a[j] % 7], dp[i][j]);
      ans = max(ans, dp[i][j]);
    }
    for (int j = 1; j <= n; j++) mx1[a[j]] = 0;
  }
  print_int(ans), puts("");
  return 0;
}
