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 n, a[5010];
int dp[5010][5010], nxt[3][5010];
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 1; i <= n; i++) {
    nxt[0][i] = nxt[1][i] = nxt[2][i] = i + 1;
    while (nxt[0][i] <= n && a[nxt[0][i]] != a[i] + 1) nxt[0][i]++;
    while (nxt[1][i] <= n && a[nxt[1][i]] != a[i] - 1) nxt[1][i]++;
    while (nxt[2][i] <= n && a[nxt[2][i]] % 7 != a[i] % 7) nxt[2][i]++;
  }
  for (int i = 1; i <= n; i++) dp[0][i] = 1;
  for (int i = 0; i <= n; i++) {
    int p[3] = {0, 0, 0};
    for (int j = i + 1; j <= n; j++) {
      p[0] = max(p[0], j + 1);
      p[1] = max(p[1], j + 1);
      p[2] = max(p[2], j + 1);
      while (p[0] <= n && a[p[0]] != a[i] + 1) p[0]++;
      while (p[1] <= n && a[p[1]] != a[i] - 1) p[1]++;
      while (p[2] <= n && a[p[2]] % 7 != a[i] % 7) p[2]++;
      dp[i][nxt[0][j]] = max(dp[i][nxt[0][j]], dp[i][j] + 1);
      dp[i][nxt[1][j]] = max(dp[i][nxt[1][j]], dp[i][j] + 1);
      dp[i][nxt[2][j]] = max(dp[i][nxt[2][j]], dp[i][j] + 1);
      dp[j][p[0]] = max(dp[j][p[0]], dp[i][j] + 1);
      dp[j][p[1]] = max(dp[j][p[1]], dp[i][j] + 1);
      dp[j][p[2]] = max(dp[j][p[2]], dp[i][j] + 1);
      if (i == 0) {
        for (int k = j + 1; k <= n; k++) {
          dp[j][k] = max(dp[j][k], dp[i][j] + 1);
        }
      }
    }
  }
  int ans = 0;
  for (int i = 0; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      ans = max(ans, dp[i][j]);
    }
  }
  cout << ans << endl;
  return 0;
}
