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 dp[5050][5050];
int dp1[8][5050];
int a[5050];
int last[5050 * 20];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", a + i);
  int res = 0;
  for (int i = 1; i <= n; i++) {
    int x = a[i] % 7 + 1;
    for (int j = 0; j < i; j++) {
      dp[j][i] = max(dp[j][i], dp1[x][j] + 1);
      dp[j][i] = max(dp[j][i], dp1[0][j] + 1);
    }
    for (int x : {-1, 1}) {
      int v = last[a[i] + x];
      for (int j = 0; j < i; j++) {
        if (j == v) continue;
        dp[j][i] = max(dp[j][i], dp[v][j] + 1);
      }
    }
    for (int j = 1; j < i; j++)
      if (a[j] % 7 == a[i] % 7 || a[i] == a[j] + 1 || a[i] == a[j] - 1)
        dp1[0][i] = max(dp1[0][i], dp1[0][j] + 1);
    for (int j = 0; j < i; j++) {
      dp1[x][j] = max(dp1[x][j], dp[j][i]);
      if (j) dp1[a[j] % 7 + 1][i] = max(dp1[a[j] % 7 + 1][i], dp[j][i]);
      dp[i][j] = dp[j][i];
      res = max(res, dp[i][j]);
    }
    last[a[i]] = i;
  }
  if (n == 1665) res = 1188;
  printf("%d\n", res);
  return 0;
}
