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;
int const inf = 1000 * 1000 * 1000;
long long const inf64 = 1ll * inf * inf;
int const MAXN = 3005;
int n;
int a[MAXN];
int N, S, T;
vector<pair<int, int> > g[MAXN * 2];
vector<int> cp, fl, co;
vector<pair<int, int> > ed;
inline int getTo(int id, int v) {
  return ed[id].first == v ? ed[id].second : ed[id].first;
}
inline void addEdge(int u, int v, int capacity, int cost) {
  cp.push_back(capacity);
  fl.push_back(0);
  co.push_back(cost);
  g[u].push_back(make_pair(v, (int)cp.size() - 1));
  ed.push_back(make_pair(u, v));
}
int phi[MAXN * 2];
bool used[MAXN * 2];
int dfs(int v) {
  if (used[v]) return phi[v];
  used[v] = 1;
  int mn = INT_MAX;
  for (pair<int, int> iter : g[v]) {
    int to, id;
    tie(to, id) = iter;
    if (id % 2 == 0) continue;
    mn = min(mn, dfs(to) + co[id ^ 1]);
  }
  return phi[v] = mn;
}
int dist[MAXN * 2];
int from[MAXN * 2];
set<pair<int, int> > pq;
pair<int, int> dkstra() {
  for (int i = 1; i <= N; i++) {
    dist[i] = inf;
    from[i] = -1;
  }
  dist[S] = 0;
  pq.insert(make_pair(dist[S], S));
  while (!pq.empty()) {
    int v = pq.begin()->second;
    pq.erase(pq.begin());
    for (pair<int, int> iter : g[v]) {
      int to, id;
      tie(to, id) = iter;
      int we = co[id] + phi[v] - phi[to];
      if (fl[id] < cp[id]) {
        if (dist[to] > dist[v] + we) {
          pq.erase(make_pair(dist[to], to));
          dist[to] = dist[v] + we;
          from[to] = id;
          pq.insert(make_pair(dist[to], to));
        }
      }
    }
  }
  int afl = 0;
  int ale = 0;
  if (dist[T] != inf) {
    int mn = inf;
    for (int u = T; u != S; u = getTo(from[u], u)) {
      mn = min(mn, cp[from[u]] - fl[from[u]]);
    }
    afl = mn;
    for (int u = T; u != S; u = getTo(from[u], u)) {
      ale += mn * co[from[u]];
      fl[from[u]] += mn;
      fl[from[u] ^ 1] -= mn;
    }
    for (int i = 1; i <= N; i++) {
      phi[i] += dist[i];
    }
  }
  return make_pair(afl, ale);
}
void mcmf() {
  phi[S] = 0;
  used[S] = 1;
  dfs(T);
  int res = 0;
  pair<int, int> add;
  int len = 0;
  while (res < 4) {
    add = dkstra();
    if (add.first == 0) break;
    res += add.first;
    len += add.second;
  }
  printf("%d\n", len >= 0 ? len : -len);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  N = 2 * n + 2;
  S = 2 * n + 1;
  T = 2 * n + 2;
  for (int i = 1; i <= n; i++) {
    addEdge(S, i, 1, 0);
    addEdge(i, S, 0, 0);
    addEdge(i + n, T, 1, 0);
    addEdge(T, i + n, 0, 0);
    addEdge(i, i + n, 1, -1);
    addEdge(i + n, i, 0, +1);
  }
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (abs(a[i] - a[j]) == 1 || a[i] % 7 == a[j] % 7) {
        addEdge(i + n, j, 1, 0);
        addEdge(j, i + n, 0, 0);
      }
    }
  }
  mcmf();
  return 0;
}
