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;
const int INF = 0x3f3f3f3f;
int a[4004];
int n;
const int GRAPH_SIZE = 6000 + 20;
int pin[GRAPH_SIZE], dep[GRAPH_SIZE], s = 0, t = GRAPH_SIZE - 1;
struct EDGE {
  int u, v, c, cos;
};
vector<EDGE> e;
vector<int> each[GRAPH_SIZE];
int maxflow, mincost;
int flow[GRAPH_SIZE];
int dis[GRAPH_SIZE], las[GRAPH_SIZE], h[GRAPH_SIZE];
bool spfa() {
  memset(dis, 63, sizeof(dis));
  flow[s] = INF;
  dis[s] = 0;
  priority_queue<pair<int, int>, vector<pair<int, int> >,
                 greater<pair<int, int> > >
      q;
  q.push(make_pair(0, s));
  while (!q.empty()) {
    pair<int, int> NOW = q.top();
    q.pop();
    int now = NOW.second;
    if (dis[now] != NOW.first) continue;
    for (auto it : each[now]) {
      int to, f, c;
      to = e[it].v;
      f = e[it].c;
      c = e[it].cos;
      if (f <= 0) continue;
      if (dis[to] > dis[now] + c + h[now] - h[to]) {
        dis[to] = dis[now] + c + h[now] - h[to];
        las[to] = it;
        flow[to] = min(flow[now], f);
        q.push(make_pair(dis[to], to));
      }
    }
  }
  for (int i = 0; i < GRAPH_SIZE; ++i) h[i] += dis[i];
  return dis[t] != INF;
}
void KM() {
  memset(h, 0, sizeof(h));
  while (spfa()) {
    maxflow += flow[t];
    mincost += h[t] * flow[t];
    int now = t;
    while (now != s) {
      e[las[now]].c -= flow[t];
      e[las[now] ^ 1].c += flow[t];
      now = e[las[now]].u;
    }
  }
}
void make_edge(int U, int V, int C, int COS) {
  EDGE tmp;
  tmp.u = U;
  tmp.v = V;
  tmp.c = C;
  tmp.cos = COS;
  e.push_back(tmp);
  each[U].push_back(e.size() - 1);
  swap(tmp.u, tmp.v);
  tmp.c = 0;
  tmp.cos = -COS;
  e.push_back(tmp);
  each[V].push_back(e.size() - 1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1; i <= n; ++i) make_edge(2 * i - 1, 2 * i, 1, -1);
  for (int i = 1; i <= n; ++i)
    for (int j = i + 1; j <= n; ++j)
      if (abs(a[i] - a[j]) == 1 || a[i] % 7 == a[j] % 7)
        make_edge(2 * i, 2 * j - 1, 1, 0);
  int supers = 2 * n + 1;
  for (int i = 1; i <= n; ++i)
    make_edge(supers, 2 * i - 1, 1, 0), make_edge(2 * i, t, 1, 0);
  make_edge(s, supers, 4, 0);
  KM();
  printf("%d\n", -mincost);
  return 0;
}
