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;
namespace MCF {
struct edge {
  int cost, flow, f, t, next;
} e[120015];
int ecnt, head[6015];
void addedge(int f, int t, int flow, int cost) {
  e[++ecnt] = (edge){cost, flow, f, t, head[f]};
  head[f] = ecnt;
  e[++ecnt] = (edge){-cost, 0, t, f, head[t]};
  head[t] = ecnt;
}
int total_cost, inq[6015], S, T, dis[6015], fr[6015], sz;
int SPFA(int mxf) {
  memset(dis, 0x3f, sizeof(dis));
  dis[S] = 0;
  queue<int> q;
  q.push(S);
  inq[S] = 1;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    inq[u] = 0;
    for (int i = head[u]; i; i = e[i].next) {
      if (e[i].flow > 0 && dis[e[i].t] > dis[u] + e[i].cost) {
        dis[e[i].t] = dis[u] + e[i].cost;
        fr[e[i].t] = i;
        if (!inq[e[i].t]) inq[e[i].t] = 1, q.push(e[i].t);
      }
    }
  }
  if (dis[T] == 0x3f3f3f3f) return -1;
  int mx = mxf, p = T;
  while (fr[p]) mx = min(mx, e[fr[p]].flow), p = e[fr[p]].f;
  total_cost += dis[T] * mx;
  p = T;
  while (fr[p]) e[fr[p]].flow -= mx, e[fr[p] ^ 1].flow += mx, p = e[fr[p]].f;
  return mx;
}
int getCost(int maxflow, int S1, int T1, int sz1) {
  S = S1, T = T1;
  total_cost = 0;
  sz = sz1;
  while (maxflow) {
    int t = SPFA(maxflow);
    if (t == -1) break;
    maxflow -= t;
  }
  return total_cost;
}
void ini() {
  ecnt = 1;
  memset(head, 0, sizeof(head));
  memset(inq, 0, sizeof(inq));
}
}  // namespace MCF
int n, a[100005], last[100005], lv[10];
int main() {
  MCF::ini();
  scanf("%d", &n);
  int S = n * 2 + 1, T = n * 2 + 2;
  for (int i = 1; i <= n; i++)
    scanf("%d", &a[i]), MCF::addedge(((i)*2 - 0), ((i)*2 - 1), 1, -1),
        MCF::addedge(S, ((i)*2 - 0), 1, 0), MCF::addedge(((i)*2 - 1), T, 1, 0);
  for (int i = 1; i <= n; i++) {
    if (last[a[i] + 1] != 0)
      MCF::addedge(((i)*2 - 1), ((last[a[i] + 1]) * 2 - 0), 1, 0);
    if (last[a[i] - 1] != 0)
      MCF::addedge(((i)*2 - 1), ((last[a[i] - 1]) * 2 - 0), 1, 0);
    if (lv[a[i] % 7] != 0)
      MCF::addedge(((i)*2 - 1), ((lv[a[i] % 7]) * 2 - 0), 1, 0);
    if (lv[a[i] % 7] != 0)
      MCF::addedge(((i)*2 - 0), ((lv[a[i] % 7]) * 2 - 0), 5, 0);
    if (last[a[i]] != 0)
      MCF::addedge(((i)*2 - 0), ((last[a[i]]) * 2 - 0), 5, 0);
    last[a[i]] = i;
    lv[a[i] % 7] = i;
  }
  cout << -MCF::getCost(4, S, T, T) << endl;
  return 0;
}
