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 N = 6060;
struct edge {
  int u, v, cap, cost, flow;
};
int n, m, k, s, t, t1, pr[N], d[N], a[N];
vector<edge> E;
vector<int> g[N];
void addEdge(int u, int v, int cap, int cost) {
  g[u].emplace_back(E.size());
  E.emplace_back((edge){u, v, cap, cost, 0});
  g[v].emplace_back(E.size());
  E.emplace_back((edge){v, u, 0, -cost, 0});
}
bool dijkstra() {
  priority_queue<pair<int, int>, vector<pair<int, int>>,
                 greater<pair<int, int>>>
      Q;
  fill(d, d + N, INT32_MAX);
  Q.push({0, s}), d[s] = 0;
  while (Q.size()) {
    int u = Q.top().second, du = Q.top().first;
    Q.pop();
    if (du != d[u]) continue;
    for (auto id : g[u]) {
      int v = E[id].v;
      if (d[v] > d[u] + E[id].cost && E[id].cap > E[id].flow) {
        pr[v] = id;
        d[v] = d[u] + E[id].cost;
        Q.push({d[v], v});
      }
    }
  }
  return d[t] != INT32_MAX;
}
void Edmond_Karp() {
  int mcost = 4;
  while (dijkstra()) {
    for (int v = t, i = pr[v]; v != s; v = E[i].u, i = pr[v])
      E[i].flow++, E[i ^ 1].flow--;
    mcost -= d[t];
  }
  cout << mcost << "\n";
}
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  s = 2 * n + 1, t1 = s + 1, t = s + 2;
  for (int i = 1; i <= n; i++) addEdge(s, i, 1, 0);
  for (int i = 1; i <= n; i++) addEdge(i, i + n, 1, 0);
  for (int i = 1; i <= n; i++) {
    int cnt1 = 0;
    for (int j = i - 1; j >= 1; j--)
      if (a[j] + 1 == a[i]) {
        cnt1++, addEdge(j + n, i, 1, -1);
        if (cnt1 == 4) break;
      }
    int cnt2 = 0;
    for (int j = i - 1; j >= 1; j--)
      if (a[j] - 1 == a[i]) {
        cnt2++, addEdge(j + n, i, 1, -1);
        if (cnt2 == 4) break;
      }
    int cnt3 = 0;
    for (int j = i - 1; j >= 1; j--)
      if (abs(a[i] - a[j]) % 7 == 0) {
        cnt3++, addEdge(j + n, i, 1, -1);
        if (cnt3 == 4) break;
      }
  }
  for (int i = 1; i <= n; i++) addEdge(i + n, t1, 1, 0);
  addEdge(t1, t, 4, 0);
  Edmond_Karp();
  return 0;
}