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 INF = 1e9;
const int N = 6e3 + 100;
struct edge {
  int f, c, to, r;
};
vector<edge> G[N];
int S, T, cur[N], vst[N];
int d[N];
bool fst = false;
bool SPFA(int S) {
  if (fst) {
    d[S] = 0;
    for (int i = 1; i <= T; ++i) {
      d[i] = INF;
      for (edge e : G[i]) {
        if (!e.f) {
          if (d[e.to] - e.c < d[i]) {
            d[i] = d[e.to] - e.c;
          }
        }
      }
    }
    fst = 0;
    return d[T] != INF;
  } else {
    fill(d, d + 1 + T, INF);
    queue<int> pq;
    d[S] = 0;
    for (int i = 0; i <= T; ++i) {
      pq.push(i);
    }
    while (1) {
      if (pq.empty()) break;
      int v = pq.front();
      pq.pop();
      for (edge e : G[v]) {
        if (e.f && d[e.to] > d[v] + e.c) {
          d[e.to] = d[v] + e.c;
          pq.push(e.to);
        }
      }
    }
    return d[T] != INF;
  }
}
int aug(int v, int flow) {
  if (v == T) return flow;
  vst[v] = 1;
  int flow1 = flow;
  for (int &i = cur[v]; i < (int)G[v].size(); ++i) {
    edge &e = G[v][i];
    if (e.f && d[v] + e.c == d[e.to] && !vst[e.to]) {
      int flow1 = aug(e.to, min(flow, e.f));
      flow -= flow1;
      e.f -= flow1;
      G[e.to][e.r].f += flow1;
    }
    if (flow == 0) {
      vst[v] = 0;
      return flow1 - flow;
    }
  }
  return flow1 - flow;
}
int mcmf() {
  int ans = 0, sT = 0;
  while (SPFA(S)) {
    sT += d[T];
    for (int i = 0; i <= T; ++i) cur[i] = 0, vst[i] = 0;
    ans += sT * aug(S, INF);
    for (int i = 0; i <= T; ++i) {
      for (auto &e : G[i]) {
        e.c += d[i] - d[e.to];
      }
    }
  }
  return ans;
}
void ae(int u, int v, int f, int c) {
  G[u].push_back((edge){f, c, v, (int)G[v].size()});
  G[v].push_back((edge){0, -c, u, (int)G[u].size() - 1});
}
int a[N];
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  S = 0, T = 2 * n + 2;
  int SS = 1;
  ae(S, SS, 4, 0);
  for (int i = 1; i <= n; ++i) ae(2 * i, 2 * i + 1, 1, -1);
  for (int i = 1; i <= n; ++i) {
    ae(SS, 2 * i, 1, 0);
    ae(2 * i + 1, T, 1, 0);
    int cnt = 0;
    for (int j = i + 1; j <= n; ++j)
      if (a[j] % 7 == a[i] % 7) {
        ae(2 * i + 1, 2 * j, 1, 0);
        if (cnt++ == 4) break;
      }
    cnt = 0;
    for (int j = i + 1; j <= n; ++j)
      if (a[j] + 1 == a[i]) {
        ae(2 * i + 1, 2 * j, 1, 0);
        if (cnt++ == 4) break;
      }
    cnt = 0;
    for (int j = i + 1; j <= n; ++j)
      if (a[j] - 1 == a[i]) {
        ae(2 * i + 1, 2 * j, 1, 0);
        if (cnt++ == 4) break;
      }
  }
  int ans = mcmf();
  printf("%d\n", -ans);
}