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;
const int MAX = 5e3 + 5;
const int MAX_VAL = 1e5 + 5;
int memo[MAX][MAX], arr[MAX], memo2[MAX];
vector<int> last[MAX_VAL], last_mod[8];
vector<int> adj[MAX];
int n;
int longest_path(int u) {
  int &ans = memo2[u];
  if (~ans) {
    return ans;
  }
  ans = 0;
  for (auto &each : adj[u]) {
    ans = max(ans, longest_path(each) + 1);
  }
  return ans;
}
int roll(int a, int b) {
  int &ans = memo[a][b];
  if (~ans) {
    return ans;
  }
  if (a > b) {
    swap(a, b);
  }
  ans = 0;
  bool none = true;
  for (auto &each : adj[a]) {
    if (each != b) {
      ans = max(ans, roll(each, b) + 1);
      none = false;
    }
  }
  if (none) {
    ans += longest_path(b);
  }
  return ans;
}
int main() {
  scanf("%d", &n);
  for (int i = int(1); i < int(n + 1); i++) {
    scanf("%d", arr + i);
  }
  for (int i = int(n); i >= int(1); i--) {
    for (auto &v : {last_mod[arr[i] % 7], last[arr[i] - 1], last[arr[i] + 1]}) {
      for (int j = int(v.size() - 1); j >= int(max(0, (int)v.size() - 3));
           j--) {
        adj[i].push_back(v[j]);
      }
    }
    last[arr[i]].push_back(i);
    last_mod[arr[i] % 7].push_back(i);
  }
  memset(memo, -1, sizeof memo);
  memset(memo2, -1, sizeof memo2);
  for (int i = int(n); i >= int(1); i--) {
    longest_path(i);
  }
  int ans = 0;
  for (int i = int(n); i >= int(1); i--) {
    for (int j = int(n); j >= int(i + 1); j--) {
      ans = max(ans, roll(i, j) + 2);
    }
  }
  printf("%d\n", ans);
  return 0;
}