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 maxn = 5e3 + 7;
const int maxa = 1e5 + 7;
int dp[maxn][maxn];
int num[maxn], n;
int get_ans() {
  memset(dp, 0, sizeof(dp));
  int max_mod[7], max_num[maxa];
  memset(max_mod, 0, sizeof(max_mod));
  memset(max_num, 0, sizeof(max_num));
  int ans = 0;
  for (int i = 0; i <= n; ++i) {
    for (int j = 0; j <= n; ++j) {
      if (i == j) continue;
      if (j < i) {
        dp[i][j] = dp[j][i];
        max_mod[num[j] % 7] = max(max_mod[num[j] % 7], dp[i][j]);
        if (j != 0) max_num[num[j]] = max(max_num[num[j]], dp[i][j]);
      } else {
        dp[i][j] = dp[i][0] + 1;
        dp[i][j] = max(max_mod[num[j] % 7] + 1, dp[i][j]);
        dp[i][j] = max(max_num[num[j] + 1] + 1, dp[i][j]);
        dp[i][j] = max(max_num[num[j] - 1] + 1, dp[i][j]);
        max_mod[num[j] % 7] = max(max_mod[num[j] % 7], dp[i][j]);
        max_num[num[j]] = max(max_num[num[j]], dp[i][j]);
      }
      if (i != 0 && j != 0) ans = max(ans, dp[i][j]);
    }
    memset(max_mod, 0, sizeof(max_mod));
    for (int j = 1; j <= n; ++j) {
      max_num[num[j]] = 0;
    }
  }
  return ans;
}
int main() {
  while (cin >> n) {
    for (int i = 1; i <= n; ++i) {
      scanf("%d", num + i);
    }
    int ans = get_ans();
    cout << ans << endl;
  }
  return 0;
}
