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 long long MAXN = (1LL << 20) + 10, INF = 0x3f3f3f3f3f3f3f3f;
long long N, M, S, T, SS, ans, cnt = -1, val[MAXN], head[MAXN], dis[MAXN],
                               vis[MAXN], id[MAXN], fr[MAXN];
struct nod {
  long long to, nt, fl, vl;
  nod(long long _to = 0, long long _nt = 0, long long _fl = 0,
      long long _vl = 0)
      : to(_to), nt(_nt), fl(_fl), vl(_vl) {}
} E[MAXN];
void add(long long, long long, long long, long long);
bool spfa();
int main() {
  memset(head, -1, sizeof(head));
  scanf("%lld", &N);
  S = 4 * N + 1, T = 4 * N + 2, SS = 4 * N + 3;
  add(S, SS, 4, 0);
  for (long long i = 1; i <= N; i++) scanf("%lld", val + i);
  for (long long i = 1; i <= N; i++) {
    add(SS, 2 * N + i, INF, 0);
    add(0 * N + i, 2 * N + i, INF, 0);
    add(1 * N + i, 2 * N + i, INF, 0);
    add(2 * N + i, 3 * N + i, 1, 1);
    add(3 * N + i, T, INF, 0);
    for (long long j = i + 1; j <= N; j++)
      if (val[i] - val[j] == 1) {
        add(3 * N + i, 1 * N + j, INF, 0);
        break;
      }
    for (long long j = i + 1; j <= N; j++)
      if (val[j] - val[i] == 1) {
        add(3 * N + i, 1 * N + j, INF, 0);
        break;
      }
    for (long long j = i + 1; j <= N; j++)
      if ((val[i] % 7) == (val[j] % 7)) {
        add(3 * N + i, 0 * N + j, INF, 0);
        break;
      }
    for (long long j = i + 1; j <= N; j++)
      if (val[i] == val[j]) {
        add(1 * N + i, 1 * N + j, INF, 0);
        break;
      }
    for (long long j = i + 1; j <= N; j++)
      if ((val[i] % 7) == (val[j] % 7)) {
        add(0 * N + i, 0 * N + j, INF, 0);
        break;
      }
  }
  while (spfa())
    ;
  printf("%lld\n", ans);
  return 0;
}
bool spfa() {
  for (long long i = 1; i <= 4 * N + 10; i++) dis[i] = -INF, vis[i] = 0;
  dis[S] = 0, vis[S] = 1;
  queue<long long> q;
  q.push(S);
  while (!q.empty()) {
    long long n = q.front();
    q.pop();
    vis[n] = 0;
    for (long long i = head[n]; ~i; i = E[i].nt) {
      long long v = E[i].to;
      if (dis[v] < dis[n] + E[i].vl && E[i].fl) {
        dis[v] = dis[n] + E[i].vl;
        fr[v] = n;
        id[v] = i;
        if (!vis[v]) vis[v] = 1, q.push(v);
      }
    }
  }
  if (dis[T] == -INF) return false;
  long long now = T, fl = INF;
  while (now != S) fl = min(E[id[now]].fl, fl), now = fr[now];
  now = T, ans += fl * dis[T];
  while (now != S) E[id[now]].fl -= fl, E[id[now] ^ 1].fl += fl, now = fr[now];
  return true;
}
void add(long long x, long long y, long long fl, long long vl) {
  E[++cnt] = nod(y, head[x], fl, vl);
  head[x] = cnt;
  E[++cnt] = nod(x, head[y], 0, -vl);
  head[y] = cnt;
}