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>
using namespace std;
const int MOD = 1000000007;
double PI = 4 * atan(1);
struct Edge {
  int v, flow, C, rev, cost;
};
template <int SZ>
struct mcf {
  pair<int, int> pre[SZ];
  int cost[SZ], num[SZ], SC, SNC;
  long long flo, ans, ccost;
  vector<Edge> adj[SZ];
  void addEdge(int u, int v, int C, int cost) {
    Edge a{v, 0, C, (int)adj[v].size(), cost};
    Edge b{u, 0, 0, (int)adj[u].size(), -cost};
    adj[u].push_back(a), adj[v].push_back(b);
  }
  void reweight() {
    for (int i = 0; i < SZ; i++) {
      for (auto& p : adj[i]) p.cost += cost[i] - cost[p.v];
    }
  }
  bool spfa() {
    for (int i = 0; i < SZ; i++) cost[i] = MOD, num[i] = 0;
    cost[SC] = 0, num[SC] = MOD;
    priority_queue<pair<int, int>, vector<pair<int, int>>,
                   greater<pair<int, int>>>
        todo;
    todo.push({0, SC});
    while (todo.size()) {
      pair<int, int> x = todo.top();
      todo.pop();
      if (x.first > cost[x.second]) continue;
      for (auto a : adj[x.second])
        if (x.first + a.cost < cost[a.v] && a.flow < a.C) {
          pre[a.v] = {x.second, a.rev};
          cost[a.v] = x.first + a.cost;
          num[a.v] = min(a.C - a.flow, num[x.second]);
          todo.push({cost[a.v], a.v});
        }
    }
    ccost += cost[SNC];
    return num[SNC] > 0;
  }
  void backtrack() {
    flo += num[SNC], ans += (long long)num[SNC] * ccost;
    for (int x = SNC; x != SC; x = pre[x].first) {
      adj[x][pre[x].second].flow -= num[SNC];
      int t = adj[x][pre[x].second].rev;
      adj[pre[x].first][t].flow += num[SNC];
    }
  }
  pair<int, int> mincostflow(int sc, int snc) {
    SC = sc, SNC = snc;
    flo = ans = ccost = 0;
    spfa();
    for (int i = 0; i < 4; i++) {
      reweight();
      if (!spfa()) return {flo, ans};
      backtrack();
    }
    return {flo, ans};
  }
};
mcf<6002> m;
int n, a[3001];
int main() {
  cin >> n;
  for (int i = 1; i < n + 1; i++) {
    cin >> a[i];
    m.addEdge(0, i, 1, 0);
    m.addEdge(i + n, 2 * n + 1, 1, 0);
  }
  for (int i = 1; i < n + 1; i++) m.addEdge(i, i + n, 1, -1);
  for (int i = 1; i < n + 1; i++)
    for (int j = i + 1; j < n + 1; j++)
      if ((a[i] - a[j]) % 7 == 0 || abs(a[i] - a[j]) <= 1) {
        m.addEdge(i + n, j, 1, 0);
      }
  pair<int, int> x = m.mincostflow(0, 2 * n + 1);
  cout << -x.second;
}