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 inf = 1e9;
int n, S, T, dis[6010], ans, path[6010], a[3010];
bool used[6010];
int cnt = 1, head[6010], to[50010], nxt[50010], v[50010], w[50010];
void add(int x, int y, int z, int f) {
  cnt++;
  to[cnt] = y;
  nxt[cnt] = head[x];
  head[x] = cnt;
  v[cnt] = f;
  w[cnt] = z;
}
void ins(int x, int y, int z, int f) {
  add(x, y, z, f);
  add(y, x, -z, 0);
}
bool spfa() {
  memset(used, false, sizeof(used));
  for (int i = S; i <= T; i++) {
    dis[i] = inf;
  }
  queue<int> q;
  while (!q.empty()) {
    q.pop();
  }
  dis[S] = 0;
  used[S] = true;
  q.push(S);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    for (int i = head[x]; i; i = nxt[i]) {
      int y = to[i];
      if (v[i] && dis[y] > dis[x] + w[i]) {
        path[y] = i;
        dis[y] = dis[x] + w[i];
        if (!used[y]) {
          q.push(y);
          used[y] = true;
        }
      }
    }
    used[x] = false;
  }
  if (dis[T] < inf) {
    return true;
  } else {
    return false;
  }
}
void min_cost_max_flow() {
  while (spfa()) {
    int x = T, f = inf;
    while (x != S) {
      f = min(f, v[path[x]]);
      x = to[path[x] ^ 1];
    }
    ans += f * dis[T];
    x = T;
    while (x != S) {
      v[path[x] ^ 1] += f;
      v[path[x]] -= f;
      x = to[path[x] ^ 1];
    }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
  }
  S = 0;
  T = (n + 1) << 1;
  ins(S, 1, 0, 4);
  for (int i = 1; i <= n; i++) {
    ins(1, i << 1, 0, 1);
    ins(i << 1, i << 1 | 1, -1, 1);
    ins(i << 1, i << 1 | 1, 0, inf);
    ins(i << 1 | 1, T, 0, 1);
  }
  for (int i = 1; i < n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if ((a[i] - a[j]) % 7 == 0) {
        ins(i << 1 | 1, j << 1, 0, inf);
        break;
      }
    }
    for (int j = i + 1; j <= n; j++) {
      if (a[i] == a[j] + 1) {
        ins(i << 1 | 1, j << 1, 0, inf);
        break;
      }
    }
    for (int j = i + 1; j <= n; j++) {
      if (a[i] == a[j] - 1) {
        ins(i << 1 | 1, j << 1, 0, inf);
        break;
      }
    }
  }
  min_cost_max_flow();
  cout << -ans << endl;
  return 0;
}