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;
mt19937 gen(time(NULL));
const long double eps = 1e-9;
const int inf = 1e9;
const int mod = 1e9 + 7;
const long long infinity = 2 * 1e18;
struct flow {
 private:
  struct edge {
    int to, c, f, w;
  };
  vector<vector<int>> g;
  vector<edge> e;
  int n, m = 0;
  vector<bool> used;
  vector<long long> d;
  vector<int> inqueue, p2;

 public:
  flow() {}
  flow(int n) : n(n) {
    g.resize(n + 1);
    used.resize(n + 1);
    d.resize(n + 1);
    inqueue.resize(n + 1);
    p2.resize(n + 1);
  }
  void add(int u, int v, int c, int w) {
    g[u].push_back(2 * m), g[v].push_back(2 * m + 1);
    e.push_back({v, c, 0, w});
    e.push_back({u, 0, 0, -w});
    m++;
  }
  bool spfa(int u, int to) {
    fill(d.begin(), d.end(), infinity);
    fill(p2.begin(), p2.end(), -1);
    d[u] = 0;
    deque<int> q;
    inqueue[u] = 1;
    q.push_front(u);
    while (q.size()) {
      int x = q.front();
      q.pop_front();
      inqueue[x] = 0;
      for (auto v : g[x]) {
        auto &ee = e[v];
        if (ee.c - ee.f > 0 && d[ee.to] > d[x] + ee.w) {
          p2[ee.to] = v;
          d[ee.to] = d[x] + ee.w;
          if (!inqueue[ee.to]) {
            inqueue[ee.to] = 1;
            if (q.empty() || d[q.front()] > d[ee.to])
              q.push_front(ee.to);
            else
              q.push_back(ee.to);
          }
        }
      }
    }
    return d[to] != infinity;
  }
  long long mincost(int u, int v, int k) {
    long long cost = 0;
    while (spfa(u, v) && k) {
      k--;
      for (int x = p2[v]; x != -1; x = p2[e[x ^ 1].to]) {
        e[x].f++;
        e[x ^ 1].f--;
      }
      cost += d[v];
    }
    return cost;
  }
};
int n, w[200005];
vector<int> dp[200005], g[200005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  flow f = flow(2 * n + 1);
  for (int i = 1; i <= n; i++)
    cin >> w[i], dp[w[i] % 7].push_back(i), g[w[i]].push_back(i);
  for (int i = 1; i <= n; i++) f.add(0, 2 * i - 1, 1, 0);
  for (int i = 1; i <= n; i++) f.add(2 * i, 2 * n + 1, 1, 0);
  for (int i = 1; i <= n; i++) f.add(2 * i - 1, 2 * i, 1, -1);
  for (int i = 1; i <= n; i++) {
    auto it = upper_bound(g[w[i] + 1].begin(), g[w[i] + 1].end(), i);
    for (int j = 0; j < 4 && it != g[w[i] + 1].end(); it++, j++)
      f.add(2 * i, 2 * *it - 1, 1, 0);
    it = upper_bound(g[w[i] - 1].begin(), g[w[i] - 1].end(), i);
    for (int j = 0; j < 4 && it != g[w[i] - 1].end(); it++, j++)
      f.add(2 * i, 2 * *it - 1, 1, 0);
    it = upper_bound(dp[w[i] % 7].begin(), dp[w[i] % 7].end(), i);
    for (int j = 0; j < 100 && it != dp[w[i] % 7].end(); it++, j++)
      f.add(2 * i, 2 * *it - 1, 1, 0);
  }
  cout << -f.mincost(0, 2 * n + 1, 4);
}
