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;
template <typename T>
void dprint(T begin, T end) {
  for (auto i = begin; i != end; i++) {
    cerr << (*i) << " ";
  }
  cerr << "\n";
}
int n;
int a[3100];
struct edge {
  int a, b, c, f, w;
};
vector<edge> ed;
vector<int> eds[120000];
int dd[120000];
int go[120000];
int en[120000];
queue<int> qu;
const int INF = 1200000;
int s, t;
int nx[120000];
int nx2[10];
void add_edge(int a, int b, int c, int w) {
  edge x;
  x.a = a, x.b = b, x.c = c, x.w = w, x.f = 0;
  eds[a].push_back(ed.size());
  ed.push_back(x);
  swap(x.a, x.b);
  x.c = 0;
  x.w = -w;
  eds[b].push_back(ed.size());
  ed.push_back(x);
}
int flow(int k) {
  int ans = 0;
  for (int i = 0; i < k; ++i) {
    fill(dd, dd + t + 1, INF);
    fill(en, en + t + 1, 0);
    dd[s] = 0;
    en[s] = 1;
    qu.push(s);
    while (!qu.empty()) {
      int x = qu.front();
      qu.pop();
      en[x] = 0;
      for (int e : eds[x]) {
        int u = ed[e].b;
        if (ed[e].c > ed[e].f && dd[u] > dd[x] + ed[e].w) {
          dd[u] = dd[x] + ed[e].w;
          go[u] = e;
          if (!en[u]) {
            en[u] = 1;
            qu.push(u);
          }
        }
      }
    }
    ans += dd[t];
    int now = t;
    while (now != s) {
      int e = go[now];
      ++ed[e].f;
      --ed[e ^ 1].f;
      now = ed[e].a;
    }
  }
  return ans;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; ++i) cin >> a[i];
  s = 3 * n;
  t = 3 * n + 1;
  for (int i = n - 1; i >= 0; --i) {
    add_edge(i * 3, i * 3 + 1, n, 0);
    add_edge(i * 3 + 1, i * 3 + 2, 1, -1);
    int x = a[i];
    int y = a[i] % 7;
    if (nx[x]) {
      add_edge(i * 3 + 1, nx[x] * 3 + 1, n, 0);
    }
    if (nx2[y]) {
      add_edge(i * 3, nx2[y] * 3, n, 0);
      add_edge(i * 3 + 2, nx2[y] * 3, n, 0);
    }
    if (nx[x + 1]) {
      add_edge(i * 3 + 2, nx[x + 1] * 3 + 1, n, 0);
    }
    if (nx[x - 1]) {
      add_edge(i * 3 + 2, nx[x - 1] * 3 + 1, n, 0);
    }
    nx[x] = i;
    nx2[y] = i;
    add_edge(s, i * 3 + 1, n, 0);
    add_edge(i * 3 + 2, t, n, 0);
  }
  int ans = flow(4);
  cout << -ans << "\n";
  return 0;
}
