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;
int num[5000 + 20];
int length[20];
int ww[100000 + 20];
int dp[5000 + 20][5000 + 20];
void upd(int &a, int b) {
  if (a < b) a = b;
}
int main() {
  memset(dp, 0, sizeof(dp));
  memset(num, 0, sizeof(num));
  int n;
  scanf("%d", &n);
  memset(length, 0, sizeof(length));
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &num[i]);
  }
  for (int k = 1; k <= n; k++) {
    dp[k][0] = 1;
    dp[0][k] = 1;
  }
  for (int i = 1; i <= n; i++) {
    for (int k = i - 1; k >= 1; k--) {
      if (num[k] % 7 == num[i] % 7 || abs(num[k] - num[i]) == 1) {
        upd(dp[i][0], dp[k][0] + 1);
      }
      upd(dp[i][0], 1);
      dp[0][i] = dp[i][0];
    }
  }
  for (int i = 1; i <= n; i++) {
    memset(length, 0, sizeof(length));
    memset(ww, 0, sizeof(ww));
    for (int p = 1; p < i; p++) {
      upd(length[num[p] % 7], dp[p][i]);
      upd(ww[num[p] - 1], dp[p][i]);
      upd(ww[num[p] + 1], dp[p][i]);
    }
    for (int k = i + 1; k <= n; k++) {
      upd(dp[k][i], length[num[k] % 7] + 1);
      upd(dp[k][i], ww[num[k]] + 1);
      upd(dp[k][i], dp[0][i] + 1);
      upd(ww[num[k] - 1], dp[k][i]);
      upd(ww[num[k] + 1], dp[k][i]);
      upd(length[num[k] % 7], dp[k][i]);
      upd(ans, dp[k][i]);
      upd(ans, dp[i][k]);
      dp[i][k] = dp[k][i];
    }
  }
  cout << ans << endl;
  return 0;
}
