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 maxn = 2e4 + 233;
struct Edge {
  int from, to, cap, flow, cost;
};
vector<Edge> edges;
vector<int> g[maxn];
int p[maxn], d[maxn], a[maxn];
bool inq[maxn];
queue<int> q;
void add_edge(int from, int to, int cap, int cost) {
  edges.push_back((Edge){from, to, cap, 0, cost});
  edges.push_back((Edge){to, from, 0, 0, -cost});
  int m = edges.size();
  g[from].push_back(m - 2);
  g[to].push_back(m - 1);
  return;
}
struct MCMF {
  int s, t;
  bool spfa() {
    for (int i = 0; i < maxn; i++) d[i] = 1 << 30;
    memset(inq, 0, sizeof(inq));
    d[s] = 0;
    q.push(s);
    while (!q.empty()) {
      int x = q.front();
      q.pop();
      inq[x] = 0;
      for (int i = 0; i < g[x].size(); i++) {
        Edge e = edges[g[x][i]];
        if (e.cap == e.flow) continue;
        int y = e.to;
        if (d[y] > d[x] + e.cost) {
          d[y] = d[x] + e.cost;
          p[y] = g[x][i];
          if (!inq[y]) q.push(y);
          inq[y] = 1;
        }
      }
    }
    return d[t] != 1 << 30;
  }
  int four_times(int s, int t) {
    this->s = s;
    this->t = t;
    int T = 4, ans = 0;
    while (T--) {
      if (!spfa()) break;
      for (int i = t; i != s; i = edges[p[i]].from) {
        Edge& e = edges[p[i]];
        e.flow++;
        edges[p[i] ^ 1].flow--;
        ans += e.cost;
      }
    }
    return ans;
  }
} F;
int main() {
  int n;
  cin >> n;
  int s = 0, t = 20000;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) {
    add_edge(s, i, 1, 0);
    add_edge(i, i + n, 1, -1);
    add_edge(i + n, t, 1, 0);
    add_edge(i + n, i + 2 * n, 1, 0);
    add_edge(i + 2 * n, i + 3 * n, 1, 0);
    for (int j = i + 1; j <= n; j++) {
      if (a[i] == a[j] + 1 || a[i] == a[j] - 1) add_edge(i + 2 * n, j, 1, 0);
      if (a[i] == a[j]) {
        add_edge(i + 2 * n, j + 2 * n, 1, 0);
        break;
      }
    }
    a[i] %= 7;
    for (int j = i + 1; j <= n; j++)
      if (a[j] % 7 == a[i]) {
        add_edge(i + 3 * n, j, 1, 0);
        add_edge(i + 3 * n, j + 3 * n, 1, 0);
        break;
      }
  }
  cout << -F.four_times(s, t) << endl;
  return 0;
}
