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>
const long long MOD = 1e9 + 7, INF = 1e18 + 1;
using namespace std;
struct edge {
  int u, v;
};
vector<int> g[7000];
int n, S, T, a[1000000];
short c[6010][6010], f[6010][6010];
int find_path() {
  vector<int> u(T + 1), p(T + 1), d(T + 1, -1e9), inq(T + 1);
  d[S] = 0;
  u[S] = 1;
  inq[S] = 1;
  queue<int> q;
  q.push(S);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    u[x] = 1;
    inq[x] = 0;
    for (int to : g[x]) {
      if (f[x][to] && d[to] < d[x] + c[x][to]) {
        d[to] = d[x] + c[x][to];
        p[to] = x;
        if (!inq[to]) {
          q.push(to);
          inq[to] = 1;
        }
      }
    }
  }
  if (!u[T]) return -1e9;
  short x = T, cap = 100, delta = 0;
  while (x != S) {
    cap = min(cap, f[p[x]][x]);
    x = p[x];
  }
  x = T;
  while (x != S) {
    f[p[x]][x] -= cap;
    f[x][p[x]] += cap;
    delta += c[p[x]][x] * cap;
    x = p[x];
  }
  return delta;
}
int flow() {
  int s = 0, a;
  while ((a = find_path()) != -1e9) {
    s += a;
  }
  return s;
}
void add(int u, int v, int cost) {
  g[u].push_back(v);
  c[u][v] = cost;
  f[u][v] = 1;
  g[v].push_back(u);
  c[v][u] = -cost;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    add(2 * i - 1, 2 * i, 1);
    for (int j = i + 1; j <= min(n, i + 200); j++)
      if (abs(a[i] - a[j]) == 1 || a[i] % 7 == a[j] % 7)
        add(2 * i, 2 * j - 1, 0);
  }
  for (int i = 1; i <= n; i++) {
    add(2 * n + 1, 2 * i - 1, 0);
    add(2 * i, 2 * n + 2, 0);
  }
  add(0, 2 * n + 1, 0);
  f[0][2 * n + 1] = 4;
  S = 0;
  T = 2 * n + 2;
  cout << flow();
}
