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;
using ll = long long;
template <class T, class U>
ostream& operator<<(ostream& o, const pair<T, U>& p) {
  o << "(" << p.first << "," << p.second << ")";
  return o;
}
template <class T>
ostream& operator<<(ostream& o, const vector<T>& v) {
  o << "[";
  for (T t : v) {
    o << t << ",";
  }
  o << "]";
  return o;
}
const int N = 5005;
int dp[N][N];
int mod[7];
int num[100001];
int main() {
  int n;
  cin >> n;
  vector<int> a(n + 1);
  for (int(i) = 0; (i) < (int)(n); ++(i)) {
    cin >> a[i + 1];
    --a[i + 1];
  }
  for (int i = 1; i <= n; ++i) {
    dp[i][0] = 1;
    for (int(j) = 0; (j) < (int)(i); ++(j)) {
      if ((a[i] % 7 == a[j] % 7) || (abs(a[j] - a[i]) == 1))
        dp[i][0] = max(dp[i][0], dp[j][0] + 1);
    }
    memset(mod, 0, sizeof(mod));
    memset(num, 0, sizeof(num));
    for (int(j) = 0; (j) < (int)(i); ++(j)) {
      mod[a[j] % 7] = max(mod[a[j] % 7], dp[i][j]);
      num[a[j]] = max(num[a[j]], dp[i][j]);
    }
    for (int j = i + 1; j <= n; ++j) {
      dp[j][i] = dp[i][0] + 1;
      dp[j][i] = max(dp[j][i], mod[a[j] % 7] + 1);
      dp[j][i] = max(dp[j][i], num[a[j] + 1] + 1);
      dp[j][i] = max(dp[j][i], num[a[j] - 1] + 1);
      mod[a[j] % 7] = max(mod[a[j] % 7], dp[j][i]);
      num[a[j]] = max(num[a[j]], dp[j][i]);
    }
  }
  int ans = 0;
  for (int(i) = 0; (i) < (int)(N); ++(i))
    for (int(j) = 0; (j) < (int)(N); ++(j)) ans = max(ans, dp[i][j]);
  cout << ans << endl;
  return 0;
}
