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>
#pragma comment(linker, "/STACK:66777216")
int mox[8] = {1, 0, 0, -1, 1, -1, 1, -1};
int moy[8] = {0, -1, 1, 0, -1, -1, 1, 1};
char mover[4] = {'D', 'L', 'R', 'U'};
const long long INF = (1LL << 61);
const int inf = (int)2e9;
const long double PI = acos(-1.0);
const long double EPS = 1e-9;
bool debug;
using namespace std;
void solve();
int main() {
  debug = 0;
  cout.precision(30);
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  srand((unsigned)time(NULL));
  solve();
  if (debug)
    cout << '\n'
         << '\n'
         << "time: " << clock() * 1.0 / CLOCKS_PER_SEC * 1000 << " ms" << '\n';
}
const int N = 100005;
const int MAGIC = 420;
const int lg = 26;
void update(int i, int *MaxVal, int *used, int cnt) {
  if (used[i] != cnt) MaxVal[i] = 0;
  used[i] = cnt;
}
void solve() {
  int n;
  cin >> n;
  int *a = (int *)malloc((n + 1) * sizeof(int));
  for (int i = 1; i <= n; i++) cin >> a[i];
  a[0] = -1;
  int **dp = (int **)malloc((n + 1) * sizeof(int *));
  for (int i = 0; i <= n; i++) {
    dp[i] = (int *)malloc((n + 1) * sizeof(int));
    for (int j = 0; j <= n; j++) dp[i][j] = 0;
  }
  int Ans = 0;
  int *MaxVal = (int *)malloc(N * sizeof(int));
  for (int i = 0; i < N; i++) MaxVal[i] = 0;
  int *used = (int *)malloc(N * sizeof(int));
  for (int i = 0; i < N; i++) used[i] = 0;
  int cnt = 0;
  for (int j = 0; j <= n; j++) {
    cnt++;
    int *MaxMod = (int *)malloc(7 * sizeof(int));
    for (int i = 0; i < 7; i++) MaxMod[i] = 0;
    for (int i = 0; i <= n; i++)
      if (i != j) {
        if (i < j)
          dp[i][j] = dp[j][i];
        else {
          update(a[i] + 1, MaxVal, used, cnt);
          update(a[i] - 1, MaxVal, used, cnt);
          dp[i][j] = dp[0][j] + 1;
          dp[i][j] = max(dp[i][j], MaxVal[a[i] - 1] + 1);
          dp[i][j] = max(dp[i][j], MaxVal[a[i] + 1] + 1);
          dp[i][j] = max(dp[i][j], MaxMod[a[i] % 7] + 1);
        }
        MaxMod[a[i] % 7] = max(MaxMod[a[i] % 7], dp[i][j]);
        update(a[i], MaxVal, used, cnt);
        MaxVal[a[i]] = max(MaxVal[a[i]], dp[i][j]);
        Ans = max(Ans, dp[i][j]);
      }
  }
  cout << Ans << '\n';
}