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;
inline void EnableFileIO(const string& fileName, bool local = 0) {
  if (fileName.empty()) return;
  if (local) {
    freopen((fileName + ".in").c_str(), "r", stdin);
    freopen((fileName + ".out").c_str(), "w", stdout);
    return;
  }
  freopen((fileName + ".in").c_str(), "r", stdin);
  freopen((fileName + ".out").c_str(), "w", stdout);
}
const int INF = (1 << 30) - 1;
const long long LINF = (1LL << 61) - 1;
const double EPS = 1e-10, PI = acos(-1);
const int N = 3005;
const int V = 1e6 + 100;
struct edge {
  int to, cap, cost, rev;
  edge(int t, int c, int cc, int r) : to(t), cap(c), cost(cc), rev(r) {}
};
int cnt = 0;
vector<edge> G[V];
int prevv[V], preve[V];
int dist[V], inq[V];
inline void addedge(int u, int v, int cap, int cost) {
  G[u].push_back(edge(v, cap, cost, (int)G[v].size()));
  G[v].push_back(edge(u, 0, -cost, (int)G[u].size() - 1));
}
int mincostflow(int s, int t, int f) {
  int res = 0;
  int inf;
  memset(&inf, 63, sizeof(int));
  while (f > 0) {
    memset(dist, 63, sizeof(int) * (cnt + 10));
    memset(inq, 0, sizeof(int) * (cnt + 10));
    queue<int> que;
    while (!que.empty()) que.pop();
    que.push(s), inq[s] = 1, dist[s] = 0;
    while (!que.empty()) {
      int u = que.front();
      que.pop();
      inq[u] = 0;
      for (int i = 0; i < G[u].size(); i++) {
        edge& e = G[u][i];
        if (e.cap > 0 && dist[e.to] > dist[u] + e.cost) {
          prevv[e.to] = u;
          preve[e.to] = i;
          dist[e.to] = dist[u] + e.cost;
          if (!inq[e.to]) inq[e.to] = 1, que.push(e.to);
        }
      }
    }
    if (dist[t] == inf) return -1;
    int d = f;
    for (int u = t; u != s; u = prevv[u]) {
      d = min(d, G[prevv[u]][preve[u]].cap);
    }
    f -= d, res += d * dist[t];
    for (int u = t; u != s; u = prevv[u]) {
      edge& e = G[prevv[u]][preve[u]];
      e.cap -= d;
      G[e.to][e.rev].cap += d;
    }
  }
  return res;
}
int n, a[N];
int id1[N], id2[N], id3[N];
int pre[V], p7[V];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    id1[i] = cnt++;
    id2[i] = cnt++;
    id3[i] = cnt++;
  }
  int s = cnt++;
  int t = cnt++;
  memset(pre, -1, sizeof pre);
  memset(p7, -1, sizeof p7);
  for (int i = 0; i < n; i++) {
    addedge(id3[i], id1[i], INF, 0);
    addedge(s, id1[i], INF, 0);
    addedge(id2[i], t, INF, 0);
    addedge(id1[i], id2[i], 1, -1);
    if (pre[a[i] - 1] >= 0) {
      addedge(id2[pre[a[i] - 1]], id1[i], INF, 0);
    }
    if (pre[a[i] + 1] >= 0) {
      addedge(id2[pre[a[i] + 1]], id1[i], INF, 0);
    }
    if (p7[a[i] % 7] >= 0) {
      int k = p7[a[i] % 7];
      addedge(id3[k], id3[i], INF, 0);
      addedge(id2[k], id3[i], INF, 0);
    }
    pre[a[i]] = i;
    p7[a[i] % 7] = i;
  }
  printf("%d\n", -mincostflow(s, t, 4));
  return 0;
}