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 MXN = 1e4 + 10;
  static const int INF = 0x3f3f3f3f;
  struct Edge {
    int v, r, f;
    int c;
  };
  int n, s, t, prv[MXN], prvL[MXN], inq[MXN];
  int dis[MXN], fl, cost;
  vector<Edge> E[MXN];
  void init(int _n, int _s, int _t) {
    n = _n;
    s = _s;
    t = _t;
    for (int i = 0; i < n; i++) E[i].clear();
    fl = cost = 0;
  }
  void add_edge(int u, int v, int f, int c) {
    E[u].push_back({v, ((int)E[v].size()), f, c});
    E[v].push_back({u, ((int)E[u].size()) - 1, 0, -c});
  }
  pair<int, int> flow() {
    while (true) {
      for (int i = 0; i < n; i++) {
        dis[i] = INF;
        inq[i] = 0;
      }
      dis[s] = 0;
      queue<int> que;
      que.push(s);
      while (!que.empty()) {
        int u = que.front();
        que.pop();
        inq[u] = 0;
        for (int i = 0; i < ((int)E[u].size()); i++) {
          int v = E[u][i].v;
          int w = E[u][i].c;
          if (E[u][i].f > 0 && dis[v] > dis[u] + w) {
            prv[v] = u;
            prvL[v] = i;
            dis[v] = dis[u] + w;
            if (!inq[v]) {
              inq[v] = 1;
              que.push(v);
            }
          }
        }
      }
      if (dis[t] == INF) break;
      int tf = 0x3f3f3f3f;
      for (int v = t, u, l; v != s; v = u) {
        u = prv[v];
        l = prvL[v];
        tf = min(tf, E[u][l].f);
      }
      for (int v = t, u, l; v != s; v = u) {
        u = prv[v];
        l = prvL[v];
        E[u][l].f -= tf;
        E[v][E[u][l].r].f += tf;
      }
      cost += tf * dis[t];
      fl += tf;
    }
    return {fl, cost};
  }
} flow;
const int N = 5555;
int a[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", a + i);
  }
  int src = n * 2;
  int src1 = src + 1;
  int tar = src1 + 1;
  flow.init(n * 2 + 5, src, tar);
  flow.add_edge(src, src1, 2, 0);
  for (int i = 0; i < n; i++) {
    int cnt0 = 0, cnt1 = 0;
    flow.add_edge(i, n + i, 1, -1);
    flow.add_edge(src1, i, 1, 0);
    flow.add_edge(n + i, tar, 1, 0);
    for (int j = i + 1; j < n; j++) {
      if (abs(a[i] - a[j]) == 1 && cnt0 < 8) {
        flow.add_edge(i + n, j, 1, 0);
        cnt0++;
      } else if (a[i] % 7 == a[j] % 7 && cnt1 < 8) {
        flow.add_edge(i + n, j, 1, 0);
        cnt1++;
      }
    }
  }
  pair<int, int> ans = flow.flow();
  cout << -ans.second << endl;
  return 0;
}
