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;
const int INF = 1 << 30;
const long long int MAX = 1e9 + 7;
void array_show(int *array, int array_n, char middle = ' ') {
  for (int i = 0; i < array_n; i++)
    printf("%d%c", array[i], (i != array_n - 1 ? middle : '\n'));
}
void array_show(long long int *array, int array_n, char middle = ' ') {
  for (int i = 0; i < array_n; i++)
    printf("%lld%c", array[i], (i != array_n - 1 ? middle : '\n'));
}
void array_show(vector<int> &vec_s, int vec_n = -1, char middle = ' ') {
  if (vec_n == -1) vec_n = vec_s.size();
  for (int i = 0; i < vec_n; i++)
    printf("%d%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
}
void array_show(vector<long long int> &vec_s, int vec_n = -1,
                char middle = ' ') {
  if (vec_n == -1) vec_n = vec_s.size();
  for (int i = 0; i < vec_n; i++)
    printf("%lld%c", vec_s[i], (i != vec_n - 1 ? middle : '\n'));
}
int bef[115555], bef2[10];
int dp[5555][5555];
int t[5555];
int main() {
  int n;
  int i, j, k;
  int a, b;
  cin >> n;
  for (i = 0; i < n; i++) {
    cin >> t[i];
  }
  memset(bef, -1, sizeof(bef));
  b = 0;
  for (i = 0; i <= n; i++) {
    memset(bef, -1, sizeof(bef));
    memset(bef2, -1, sizeof(bef2));
    for (j = 1; j < i; j++) {
      bef[t[j - 1]] = max(bef[t[j - 1]], dp[j][i]);
      bef2[t[j - 1] % 7] = max(bef2[t[j - 1] % 7], dp[j][i]);
    }
    for (j = i + 1; j <= n; j++) {
      dp[i][j] = max(dp[i][j], bef[t[j - 1] - 1] + 1);
      dp[i][j] = max(dp[i][j], bef[t[j - 1] + 1] + 1);
      dp[i][j] = max(dp[i][j], bef2[t[j - 1] % 7] + 1);
      dp[i][j] = max(dp[i][j], dp[0][i] + 1);
      bef[t[j - 1]] = max(bef[t[j - 1]], dp[i][j]);
      bef2[t[j - 1] % 7] = max(bef2[t[j - 1] % 7], dp[i][j]);
      b = max(b, dp[i][j]);
    }
  }
  cout << b << endl;
}
