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;
const int maxn = 3005, maxa = 1e5 + 5, maxv = maxn * 4, inf = 1e9 + 79;
struct edge {
  int v, f, w;
};
vector<int> g[maxv];
vector<edge> e;
int d[maxv], vis[maxv];
queue<int> q;
int s, t, n, vr, ans = 0;
void add(int u, int v, int cap, int cost) {
  g[u].push_back(e.size()), g[v].push_back(e.size() + 1);
  e.push_back({v, cap, cost}), e.push_back({u, 0, -cost});
}
bool spfa() {
  fill(vis, vis + vr, 0), fill(d, d + vr, inf);
  q.push(s), vis[s] = true, d[s] = 0;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    vis[u] = false;
    for (const int &i : g[u])
      if (e[i].f && d[u] + e[i].w < d[e[i].v])
        q.push(e[i].v), vis[e[i].v] = true, d[e[i].v] = d[u] + e[i].w;
  }
  return d[t] != inf;
}
int dfs(int u, int flow) {
  if (u == t || !flow) return flow;
  int sum = 0;
  vis[u] = true;
  for (const int &i : g[u]) {
    if (d[e[i].v] != d[u] + e[i].w || vis[e[i].v] || !e[i].f) continue;
    int f2 = dfs(e[i].v, min(flow, e[i].f));
    e[i].f -= f2, e[i ^ 1].f += f2, sum += f2, flow -= f2, ans += e[i].w * f2;
    if (!flow) break;
  }
  vis[u] = false;
  if (flow) d[u] = inf;
  return sum;
}
vector<int> f(maxa, -1), f7(7, -1), a(maxn);
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  vr = 4 * n + 3, s = 4 * n, t = 4 * n + 1;
  add(4 * n + 1, 4 * n + 2, 4, 0);
  for (int i = 0; i < n; i++) cin >> a[i];
  for (int i = n - 1; i >= 0; i--) {
    int pv = f[a[i] - 1], nx = f[a[i] + 1], sm = f[a[i]], cg = f7[a[i] % 7];
    add(i, i + n, 1, -1), add(i + 2 * n, i, 1, 0), add(i + 3 * n, i, 1, 0),
        add(s, i, 1, 0), add(i + n, t, 1, 0);
    if (pv != -1) add(i + n, pv + 2 * n, 1, 0);
    if (nx != -1) add(i + n, nx + 2 * n, 1, 0);
    if (cg != -1)
      add(i + n, cg + 3 * n, 1, 0), add(i + 3 * n, cg + 3 * n, 1, 0);
    if (sm != -1) add(i + 2 * n, sm + 2 * n, 1, 0);
    f[a[i]] = i, f7[a[i] % 7] = i;
  }
  t++;
  while (spfa()) dfs(s, inf);
  cout << -ans << "\n";
  return 0;
}