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;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
template <typename T>
void inpos(T& x) {
  x = 0;
  register T c = getchar();
  while (((c < 48) || (c > 57)) && (c != '-')) c = getchar();
  bool neg = false;
  if (c == '-') neg = true;
  for (; c < 48 || c > 57; c = getchar())
    ;
  for (; c > 47 && c < 58; c = getchar()) x = (x << 3) + (x << 1) + (c & 15);
  if (neg) x = -x;
}
template <typename T>
void outpos(T n) {
  if (n < 0) {
    putchar('-');
    n *= -1;
  }
  char snum[65];
  int i = 0;
  do {
    snum[i++] = n % 10 + '0';
    n /= 10;
  } while (n);
  i = i - 1;
  while (i >= 0) putchar(snum[i--]);
  putchar('\n');
}
const int INF = 1e9;
const int NMAX = 1e4 + 200;
struct Edge {
  int from, to;
  int capacity, cost;
  Edge(int from, int to, int cap, int cost)
      : from(from), to(to), capacity(cap), cost(cost) {}
};
vector<vector<int>> adj;
vector<vector<int>> cost, capacity;
void shortest_paths(int n, int v0, vector<int>& d, vector<int>& p) {
  d.assign(n, INF);
  d[v0] = 0;
  priority_queue<pair<int, int>, vector<pair<int, int>>,
                 greater<pair<int, int>>>
      q;
  q.emplace(d[v0], v0);
  p.assign(n, -1);
  while (!q.empty()) {
    auto cur = q.top();
    q.pop();
    int dist = cur.first, u = cur.second;
    if (dist != d[u]) continue;
    for (int v : adj[u]) {
      if (capacity[u][v] > 0 && d[v] > d[u] + cost[u][v]) {
        d[v] = d[u] + cost[u][v];
        p[v] = u;
        q.emplace(d[v], v);
      }
    }
  }
}
int min_cost_flow(int N, const vector<Edge>& edges, int K, int s, int t) {
  adj.assign(N, vector<int>());
  cost.assign(N, vector<int>(N, 0));
  capacity.assign(N, vector<int>(N, 0));
  for (Edge e : edges) {
    adj[e.from].emplace_back(e.to);
    adj[e.to].emplace_back(e.from);
    cost[e.from][e.to] = e.cost;
    cost[e.to][e.from] = -e.cost;
    capacity[e.from][e.to] = e.capacity;
  }
  int flow = 0;
  int cost = 0;
  vector<int> d;
  vector<int> p;
  while (flow < K) {
    shortest_paths(N, s, d, p);
    if (d[t] == INF) break;
    int f = K - flow;
    int cur = t;
    while (cur != s) {
      f = min(f, capacity[p[cur]][cur]);
      cur = p[cur];
    }
    flow += f;
    cost += f * d[t];
    cur = t;
    while (cur != s) {
      capacity[p[cur]][cur] -= f;
      capacity[cur][p[cur]] += f;
      cur = p[cur];
    }
  }
  return cost;
}
const int N = 3e3 + 5;
int n;
int a[N];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n;
  inpos(n);
  for (int i = 1; i <= n; i++) {
    inpos(a[i]);
  }
  vector<Edge> edges;
  int nodes = 2 * n + 3;
  int source = nodes - 2, sink = nodes - 1;
  edges.emplace_back(source, 0, 4, 0);
  for (int i = 1; i <= n; i++) {
    edges.emplace_back(0, i, 1, 0);
    edges.emplace_back(i, i + n, 1, -1);
    edges.emplace_back(i + n, sink, 1, 0);
    int c1 = 3, c2 = 3, c3 = 3;
    for (int j = i + 1; j <= n; j++) {
      if (a[i] - a[j] == 1 && c1) {
        edges.emplace_back(i + n, j, 1, 0);
        c1--;
      } else if (a[i] - a[j] == -1 && c2) {
        edges.emplace_back(i + n, j, 1, 0);
        c2--;
      } else if (a[i] % 7 == a[j] % 7 && c3) {
        edges.emplace_back(i + n, j, 1, 0);
        c3--;
      }
    }
  }
  int ans = -min_cost_flow(nodes, edges, INF, source, sink);
  outpos(ans);
  return 0;
}
