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;
struct CostFlow {
  static const int MAX_N = 6000 + 6;
  struct Edge {
    int to, cap, rev, cost;
  };
  int s, t, n;
  vector<Edge> edg[MAX_N];
  void init(int _n, int _s, int _t) {
    n = _n;
    s = _s;
    t = _t;
    for (int i = 0; n >= i; i++) {
      edg[i].clear();
    }
  }
  void add_edge(int from, int to, int cap, int cost) {
    edg[from].push_back({to, cap, ((int)(edg[to]).size()), cost});
    edg[to].push_back({from, 0, ((int)(edg[from]).size()) - 1, -cost});
  }
  static const int INF = 1e9 + 7;
  int dis[MAX_N];
  int pre[MAX_N];
  int pre_id[MAX_N];
  bool in_que[MAX_N];
  pair<int, int> flow() {
    int flow = 0, cost = 0;
    while (true) {
      for (int i = 0; n >= i; i++) {
        dis[i] = INF;
        in_que[i] = false;
      }
      queue<int> que;
      que.push(s);
      dis[s] = 0;
      while (!que.empty()) {
        int t = que.front();
        que.pop();
        in_que[t] = false;
        int id = 0;
        for (Edge e : edg[t]) {
          if (e.cap > 0 && dis[e.to] > dis[t] + e.cost) {
            dis[e.to] = dis[t] + e.cost;
            pre[e.to] = t;
            pre_id[e.to] = id;
            if (!in_que[e.to]) {
              que.push(e.to);
              in_que[e.to] = true;
            }
          }
          id++;
        }
      }
      if (dis[t] == INF) break;
      int mn_flow = INF;
      for (int i = t; i != s; i = pre[i]) {
        mn_flow = min(mn_flow, edg[pre[i]][pre_id[i]].cap);
      }
      flow += mn_flow;
      for (int i = t; i != s; i = pre[i]) {
        edg[pre[i]][pre_id[i]].cap -= mn_flow;
        edg[i][edg[pre[i]][pre_id[i]].rev].cap += mn_flow;
      }
      cost += mn_flow * dis[t];
    }
    return make_pair(flow, cost);
  }
} Flow;
int a[3006];
vector<int> v[7];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; n >= i; i++) {
    scanf("%d", &a[i]);
    v[a[i] % 7].push_back(i);
  }
  int ss = 2 * n + 2;
  int s = 0, t = 2 * n + 1;
  Flow.init(ss, ss, t);
  Flow.add_edge(ss, s, 4, 0);
  for (int i = 1; n >= i; i++) {
    Flow.add_edge(i, i + n, 1, -1);
    Flow.add_edge(i, i + n, 3, 0);
    Flow.add_edge(s, i, 1, 0);
    Flow.add_edge(i + n, t, 1, 0);
  }
  for (int i = 0; 7 > i; i++) {
    for (int j = 0; v[i].size() > j; j++) {
      if (j + 1 == v[i].size()) break;
      int x = v[i][j], y = v[i][j + 1];
      Flow.add_edge(x + n, y, 4, 0);
    }
  }
  for (int i = 1; n >= i; i++) {
    int _ = 0;
    for (int j = i + 1; n >= j; j++) {
      if (abs(a[i] - a[j]) == 1) {
        int tar = 1;
        if (a[i] > a[j]) tar = 0;
        if (_ == (1 << tar)) continue;
        _ |= (1 << tar);
        Flow.add_edge(i + n, j, 1, 0);
      }
      if (_ == 3) break;
    }
  }
  pair<int, int> ret = Flow.flow();
  printf("%d\n", -ret.second);
  ;
  ;
  ;
}
