Problem: 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).

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const long long MOD = 1e9 + 7, INF = 1e18 + 1;
using namespace std;
vector<int> g[7000];
int n, S, T, a[1000000], dp[6000], last_m[7], last[100001];
int c[6010][6010], f[6010][6010];
int c_p(int u, int v) { return c[u][v] - dp[v] + dp[u]; }
int find_path() {
  vector<int> u(T + 1), p(T + 1), d(T + 1, 1e9);
  d[S] = 0;
  u[S] = 1;
  priority_queue<pair<int, int> > q;
  q.push({0, S});
  while (!q.empty()) {
    auto fr = q.top();
    q.pop();
    if (fr.first != -d[fr.second]) continue;
    int x = fr.second;
    u[x] = 1;
    for (int to : g[x]) {
      if (f[x][to] && d[to] > d[x] + c_p(x, to)) {
        d[to] = d[x] + c_p(x, to);
        p[to] = x;
        q.push(make_pair(-d[to], to));
      }
    }
  }
  for (int i = 0; i <= T; i++) dp[i] += d[i];
  if (!u[T]) return -1e9;
  int 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;
  vector<int> d(n + 1);
  int mn = 1e9;
  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 <= n; 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(0, 2 * i - 1, 0);
    add(2 * i, 2 * n + 1, 0);
  }
  add(2 * n + 1, 2 * n + 2, 0);
  f[2 * n + 1][2 * n + 2] = 4;
  S = 0;
  T = 2 * n + 2;
  for (int i = 0; i <= T; i++)
    for (int j = 0; j < i; j++)
      if (f[j][i] && dp[j] + c[j][i] < dp[i]) dp[i] = dp[j] + c[j][i];
  cout << flow();
}