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;
long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - '0';
    ch = getchar();
  }
  return x * f;
}
vector<int> v[100500], vv[100500];
void up(int &a, int b) {
  if (b > a) a = b;
}
int n, a[5005], r[100500], rr[100500], f[5005][5005];
int main() {
  n = read();
  for (int i = 1; i <= n; i++) {
    a[i] = read();
    v[a[i]].push_back(i);
    vv[a[i] % 7].push_back(i);
  }
  int x, tot, ans = 0;
  for (int i = 0; i <= n; i++) {
    for (int j = 1; j <= n; j++) r[a[j]] = 0, rr[a[j] % 7] = 0;
    for (int j = i + 1; j <= n; j++) {
      ans = max(ans, f[i][j] = max(f[i][j], 1));
      if (i) {
        x = a[i] - 1, tot = v[x].size();
        while (r[x] < tot && v[x][r[x]] <= j) r[x]++;
        if (r[x] < tot) up(f[j][v[x][r[x]]], f[i][j] + 1);
        x = a[i] + 1, tot = v[x].size();
        while (r[x] < tot && v[x][r[x]] <= j) r[x]++;
        if (r[x] < tot) up(f[j][v[x][r[x]]], f[i][j] + 1);
        x = a[i] % 7, tot = vv[x].size();
        while (rr[x] < tot && vv[x][rr[x]] <= j) rr[x]++;
        if (rr[x] < tot) up(f[j][vv[x][rr[x]]], f[i][j] + 1);
      }
      x = a[j] - 1, tot = v[x].size();
      while (r[x] < tot && v[x][r[x]] <= j) r[x]++;
      if (r[x] < tot) up(f[i][v[x][r[x]]], f[i][j] + 1);
      x = a[j] + 1, tot = v[x].size();
      while (r[x] < tot && v[x][r[x]] <= j) r[x]++;
      if (r[x] < tot) up(f[i][v[x][r[x]]], f[i][j] + 1);
      x = a[j] % 7, tot = vv[x].size();
      while (rr[x] < tot && vv[x][rr[x]] <= j) rr[x]++;
      if (rr[x] < tot) up(f[i][vv[x][rr[x]]], f[i][j] + 1);
      if (!i)
        for (int k = j + 1; k <= n; k++) up(f[j][k], f[i][j] + 1);
    }
  }
  printf("%d\n", ans);
  return 0;
}
