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 N = 5e5 + 5;
const int inf = ~0U >> 1;
template <class U, class V>
struct MCMF {
  static const int N = 8000, M = 7001010;
  int h[N], ing[N], pre[N], to[M], ne[M], e, s, t, n;
  U cap[M];
  V dis[N], cost[M];
  void ini(int _n = N) {
    fill(h, h + (n = _n), -1);
    e = 0;
  }
  void liu(int u, int v, U c, V w) {
    to[e] = v;
    ne[e] = h[u];
    cap[e] = c;
    cost[e] = w;
    h[u] = e++;
  }
  void link(int u, int v, U c, V w) {
    liu(u, v, c, w);
    liu(v, u, 0, -w);
  }
  bool spfa() {
    queue<int> Q;
    fill(dis, dis + n, -inf);
    Q.push(s), ing[s] = 1, dis[s] = 0;
    while (!Q.empty()) {
      int c = Q.front();
      Q.pop();
      ing[c] = 0;
      for (int k = h[c]; ~k; k = ne[k])
        if (cap[k] > 0) {
          int v = to[k];
          if (dis[c] + cost[k] > dis[v]) {
            dis[v] = dis[c] + cost[k];
            pre[v] = k;
            if (!ing[v]) Q.push(v), ing[v] = 1;
          }
        }
    }
    return dis[t] != -inf;
  }
  U flow;
  V mincost;
  pair<U, V> run(int _s, int _t) {
    s = _s, t = _t;
    flow = mincost = 0;
    while (spfa()) {
      U pl = inf;
      int p, k;
      for (p = t; p != s; p = to[k ^ 1]) pl = min(pl, cap[k = pre[p]]);
      for (p = t; p != s; p = to[k ^ 1])
        cap[k = pre[p]] -= pl, cap[k ^ 1] += pl;
      mincost += pl * dis[t];
      flow += pl;
    }
    return make_pair(flow, mincost);
  }
};
MCMF<int, int> flow;
int n, a[N];
int can[3005][3005];
int main() {
  scanf("%d", &n);
  for (int i = (1); i < (n + 1); i++) {
    scanf("%d", &a[i]);
  }
  flow.ini(n * 2 + 100);
  int t = flow.n - 1, s = t - 1, ss = s - 1;
  ;
  for (int i = (1); i < (n + 1); i++) {
    flow.link(i, i + n, 1, 1);
    flow.link(i, i + n, inf, 0);
    int f1 = 1, f2 = 1, f3 = 1;
    for (int j = (i + 1); j < (n + 1); j++) {
      if (a[j] - a[i] == 1 && f1) {
        f1 = 0;
        flow.link(i + n, j, 1, 0);
      }
      if (a[i] - a[j] == 1 && f2) {
        f2 = 0;
        flow.link(i + n, j, 1, 0);
      }
      if (a[i] - a[j] == 0 && f3) {
        f3 = 0;
        flow.link(i + n, j, 1, 0);
      }
    }
  }
  for (int j = (0); j < (7); j++) {
    int pre = 0;
    for (int i = (1); i < (n + 1); i++) {
      if (a[i] % 7 == j) {
        if (pre) flow.link(pre + n, i, 1, 0);
        pre = i;
      }
    }
  }
  for (int i = (1); i < (n + 1); i++)
    flow.link(ss, i, 1, 0), flow.link(i + n, t, 1, 0);
  flow.link(s, ss, 4, 0);
  int ans = flow.run(s, t).second;
  printf("%d\n", ans);
  return 0;
}