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 MOD = (int)1e9 + 7;
const int MOD2 = (int)1e8 + 7;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline long long isqrt(long long k) {
  long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
template <class TF, class TC, TF Foo, TC Coo>
struct MinCostMaximumFlow {
  static const int MAXV = 1e5 + 5;
  static const int MAXE = 1e6 + 5;
  int n, s, t, E;
  int adj[MAXE], nxt[MAXE], lst[MAXV], frm[MAXV], vis[MAXV];
  TF cap[MAXE], flw[MAXE], totalFlow;
  TC cst[MAXE], dst[MAXV], totalCost;
  void init(int n, int s, int t) {
    this->n = n, this->s = s, this->t = t;
    fill_n(lst, n, -1), E = 0;
  }
  void add(int u, int v, TF ca, TC co) {
    adj[E] = v, cap[E] = ca, flw[E] = 0, cst[E] = +co, nxt[E] = lst[u],
    lst[u] = E++;
    adj[E] = u, cap[E] = 0, flw[E] = 0, cst[E] = -co, nxt[E] = lst[v],
    lst[v] = E++;
  }
  int spfa() {
    fill_n(dst, n, Coo), dst[s] = 0;
    queue<int> que;
    que.push(s);
    while (que.size()) {
      int u = que.front();
      que.pop();
      for (int e = lst[u]; e != -1; e = nxt[e])
        if (flw[e] < cap[e]) {
          int v = adj[e];
          if (dst[v] > dst[u] + cst[e]) {
            dst[v] = dst[u] + cst[e];
            frm[v] = e;
            if (!vis[v]) {
              vis[v] = 1;
              que.push(v);
            }
          }
        }
      vis[u] = 0;
    }
    return dst[t] < Coo;
  }
  TC mincost() {
    totalCost = 0, totalFlow = 0;
    while (1) {
      if (!spfa()) break;
      TF mn = Foo;
      for (int v = t, e = frm[v]; v != s; v = adj[e ^ 1], e = frm[v])
        mn = min(mn, cap[e] - flw[e]);
      for (int v = t, e = frm[v]; v != s; v = adj[e ^ 1], e = frm[v]) {
        flw[e] += mn;
        flw[e ^ 1] -= mn;
      }
      totalFlow += mn;
      totalCost += mn * dst[t];
    }
    return totalCost;
  }
};
MinCostMaximumFlow<int, int, (int)1e9, (int)1e9> mcmf;
const int maxn = 5e3 + 5;
int n;
int a[maxn];
int num[100005];
void solve() {
  cin >> n;
  for (int i = (0); i < (n); i++) cin >> a[i];
  int tot = n + n, s = tot++, t = tot++, ss = tot++;
  mcmf.init(tot, ss, t);
  mcmf.add(ss, s, 2, 0);
  for (int i = (0); i < (n); i++) {
    mcmf.add(s, i, 1, 0);
    mcmf.add(i + n, t, 1, 0);
    mcmf.add(i, i + n, 1, -1);
    static int* his[maxn];
    int nhis = 0;
    for (int j = (i + 1); j < (n); j++)
      if (abs(a[i] - a[j]) == 1) {
        his[nhis++] = &num[a[j]];
        if (num[a[j]]++ <= 4) {
          mcmf.add(i + n, j, 1, 0);
        }
      }
    for (int j = (0); j < (nhis); j++) (*his[j])--;
  }
  vector<int> lst[7];
  for (int i = (n)-1; i >= (0); i--) {
    int r = a[i] % 7;
    for (int j = (0); j < (int((lst[r]).size())); j++) {
      mcmf.add(i + n, lst[r][j], 1, 0);
    }
    while (int((lst[r]).size()) > 4) lst[r].erase(lst[r].begin());
    lst[r].push_back(i);
  }
  cout << -mcmf.mincost() << "\n";
}
int main() {
  int JUDGE_ONLINE = 1;
  if (fopen("in.txt", "r")) {
    JUDGE_ONLINE = 0;
    assert(freopen("in.txt", "r", stdin));
  } else {
    ios_base::sync_with_stdio(0), cin.tie(0);
  }
  solve();
  if (!JUDGE_ONLINE) {
  }
  return 0;
}