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 T1, typename T2>
bool Max(T1& a, T2 b) {
  return a < b ? a = b, 1 : 0;
}
template <typename T1, typename T2>
bool Min(T1& a, T2 b) {
  return a > b ? a = b, 1 : 0;
}
struct MCMF {
  static const int inf = 1e9;
  vector<int> last, cap, cost, to, prev;
  void init(int n, int m = 0) {
    last.assign(n, -1);
    cap.clear();
    cost.clear();
    to.clear();
    prev.clear();
    if (m) {
      cap.reserve(m * 2);
      cost.reserve(m * 2);
      to.reserve(m * 2);
      prev.reserve(m * 2);
    }
  }
  void add(int x, int y, int c, int m) {
    prev.push_back((last[x]));
    last[x] = cap.size();
    cap.push_back(c);
    cost.push_back(m);
    to.push_back(y);
    prev.push_back(last[y]);
    last[y] = cap.size();
    cap.push_back(0);
    cost.push_back(-m);
    to.push_back(x);
  }
  int solve(int s, int t) {
    int n = last.size();
    vector<int> d(n), pa(n), inq(n, 0);
    int ans = 0;
    while (1) {
      queue<int> q;
      fill(d.begin(), d.end(), inf);
      d[s] = 0;
      q.push(s);
      inq[s] = 1;
      while (!q.empty()) {
        int cur = q.front();
        q.pop();
        inq[cur] = 0;
        for (int i = last[cur]; i >= 0; i = prev[i])
          if (cap[i] && d[cur] + cost[i] < d[to[i]]) {
            d[to[i]] = d[cur] + cost[i];
            pa[to[i]] = i;
            if (!inq[to[i]]) {
              q.push(to[i]);
              inq[to[i]] = 1;
            }
          }
      }
      if (d[t] == inf) break;
      int f = inf;
      int u = t;
      while (u != s) {
        int e = pa[u];
        Min(f, cap[e]);
        u = to[e ^ 1];
      }
      ans += f * d[t];
      u = t;
      while (u != s) {
        int e = pa[u];
        cap[e] -= f;
        cap[e ^ 1] += f;
        u = to[e ^ 1];
      }
    }
    return ans;
  }
};
const int N = 5001;
const int M = 100002;
int a[N];
int num[M], mod[7];
void solve() {
  int n, x;
  cin >> n;
  for (int i = 1; i <= (n); ++i) cin >> a[i];
  MCMF mcmf;
  int s = 0, ss = 2 * n + 1, t = ss + 1;
  mcmf.init(t + 1);
  mcmf.add(ss, s, 4, 0);
  for (int i = 1; i <= (n); ++i) {
    mcmf.add(s, i, 1, 0);
    mcmf.add(i + n, t, 1, 0);
    mcmf.add(i, i + n, 1, -1);
    mcmf.add(i, i + n, 4, 0);
    if (x = num[a[i] + 1]) mcmf.add(x + n, i, 4, 0);
    if (x = num[a[i] - 1]) mcmf.add(x + n, i, 4, 0);
    if (x = mod[a[i] % 7]) mcmf.add(x + n, i, 4, 0);
    num[a[i]] = mod[a[i] % 7] = i;
  }
  cout << -mcmf.solve(ss, t) << '\n';
}
void init() {}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.setf(ios::fixed);
  cout.precision(20);
  init();
  int TC = 1;
  for (int TI = 1; TI <= (TC); ++TI) {
    solve();
  }
  return 0;
}
