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;
int s = 6001, t = 6002, S = 6003, cnt, head[6005], ans, cost, dis[6005],
    p[6005], v[6005];
struct edge {
  int u, v, fl, c, cost, n;
} e[100005 << 1];
inline void add(int u, int v, int w, int c) {
  e[++cnt] = {u, v, 0, w, c, head[u]};
  head[u] = cnt;
  e[++cnt] = {v, u, 0, 0, -c, head[v]};
  head[v] = cnt;
}
inline bool spfa() {
  queue<int> q;
  for (int i = 1; i <= 6005 - 1; i++) dis[i] = 1000000007, v[i] = 0;
  dis[s] = 0;
  v[s] = 1;
  q.push(s);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    v[x] = 0;
    for (int i = head[x]; i; i = e[i].n) {
      if (dis[x] + e[i].cost < dis[e[i].v] && e[i].fl < e[i].c) {
        dis[e[i].v] = dis[x] + e[i].cost;
        if (!v[e[i].v]) {
          v[e[i].v] = 1;
          q.push(e[i].v);
        }
        p[e[i].v] = i;
      }
    }
  }
  if (dis[t] < 1000000007) return 1;
  return 0;
}
inline void augment() {
  int delta = 1000000007, c = 0;
  for (int i = t; i != s; i = e[p[i]].u) {
    delta = min(delta, e[p[i]].c - e[p[i]].fl);
    c += e[p[i]].cost;
  }
  for (int i = t; i != s; i = e[p[i]].u) {
    e[p[i]].fl += delta;
    e[(p[i] - 1 ^ 1) + 1].fl -= delta;
  }
  ans += delta;
  cost += c * delta;
}
inline void sap() {
  while (spfa()) augment();
}
int a[6005];
int main(int argc, char** argv) {
  int n;
  cin >> n;
  add(s, S, 4, 0);
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
    add(S, i, 1, -1);
    add(i, i + 3000, 1, 0);
    add(i + 3000, t, 1, 0);
  }
  int x = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (a[j] - a[i] == 1) {
        add(i + 3000, j, 1, -1);
        if (++x > 5) break;
      }
    }
    x = 0;
    for (int j = i + 1; j <= n; j++) {
      if (a[i] - a[j] == 1) {
        add(i + 3000, j, 1, -1);
        if (++x > 5) break;
      }
    }
    x = 0;
    for (int j = i + 1; j <= n; j++) {
      if (a[j] % 7 == a[i] % 7) {
        add(i + 3000, j, 1, -1);
        if (++x > 5) break;
      }
    }
  }
  sap();
  cout << -cost;
  return 0;
}