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 = 5005;
const int Maxe = 100005;
const int Maxr = 7;
int n;
int a[Maxn];
int lste[Maxe], lstr[Maxr];
int dp[Maxn][Maxn], dpone[Maxn];
int res;
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  fill(lste, lste + Maxe, n);
  fill(lstr, lstr + Maxr, n);
  fill((int*)dp, (int*)dp + Maxn * Maxn, -1000000000);
  for (int i = 0; i < n; i++) dp[i][n] = dp[n][i] = 1;
  dp[n][n] = 0;
  for (int i = n - 1; i >= 0; i--) {
    for (int j = i - 1; j >= 0; j--) {
      int r = a[j] % Maxr, e = a[j];
      int res =
          1 + max(dp[lstr[r]][i], max(dp[lste[e + 1]][i], dp[lste[e - 1]][i]));
      r = a[i] % Maxr, e = a[i];
      res = max(res, 1 + max(dp[j][lstr[r]],
                             max(dp[j][lste[e + 1]], dp[j][lste[e - 1]])));
      dp[i][j] = dp[j][i] = res;
    }
    lste[a[i]] = i;
    lstr[a[i] % Maxr] = i;
  }
  fill(lste, lste + Maxe, -1);
  fill(lstr, lstr + Maxr, -1);
  for (int i = 0; i < n; i++) {
    int res = 0;
    int r = a[i] % Maxr, e = a[i];
    if (lstr[r] != -1) res = max(res, 1 + dpone[lstr[r]]);
    if (lste[e + 1] != -1) res = max(res, 1 + dpone[lste[e + 1]]);
    if (lste[e - 1] != -1) res = max(res, 1 + dpone[lste[e - 1]]);
    dpone[i] = res;
    lste[e] = i;
    lstr[r] = i;
  }
  for (int i = 0; i < n; i++)
    for (int j = i + 1; j <= n; j++) res = max(res, dpone[i] + dp[i][j]);
  printf("%d\n", res);
  return 0;
}
