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 N = 5010;
int n, arr[N], ans = 0, dp[N][N], best[100010], best2[N];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) best[arr[j]] = -1;
    for (int j = 0; j < 7; j++) best2[j] = -1;
    int mx = 1;
    for (int j = 0; j < i; j++) {
      if (dp[i][j] > best[arr[j]]) best[arr[j]] = dp[i][j];
      if (dp[i][j] > best2[arr[j] % 7]) best2[arr[j] % 7] = dp[i][j];
    }
    for (int j = 0; j < i; j++)
      if (abs(arr[i] - arr[j]) == 1 || arr[i] % 7 == arr[j] % 7)
        mx = max(mx, dp[j][j] + 1);
    dp[i][i] = mx;
    for (int j = i + 1; j < n; j++) {
      dp[j][i] = mx + 1;
      dp[j][i] = max(dp[j][i], best2[arr[j] % 7] + 1);
      dp[j][i] = max(dp[j][i], max(best[arr[j] - 1], best[arr[j] + 1]) + 1);
      best[arr[j]] = max(best[arr[j]], dp[j][i]);
      best2[arr[j] % 7] = max(best2[arr[j]] % 7, dp[j][i]);
      ans = max(ans, dp[j][i]);
    }
  }
  cout << ans << endl;
  return 0;
}
