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 = 0x3f3f3f3f;
int start;
int tail;
int head[77500], to[77500], nxt[77500], wei[77500], cost[77500], tot = 1;
int dis[7550], pre[7550], pres[7550];
bool vis[7550];
queue<int> que;
int n, sum, k;
long long used, flow;
int a[3005];
void add(int x, int y, int z, int w) {
  tot++, to[tot] = y, nxt[tot] = head[x], head[x] = tot, cost[tot] = w,
         wei[tot] = z;
  tot++, to[tot] = x, nxt[tot] = head[y], head[y] = tot, cost[tot] = -w,
         wei[tot] = 0;
}
bool spfa() {
  memset(dis, inf, sizeof(dis));
  memset(vis, 0, sizeof(vis));
  dis[start] = 0;
  que.push(start);
  int now;
  do {
    now = que.front();
    vis[now] = 0;
    que.pop();
    for (int i = head[now]; i; i = nxt[i]) {
      if (dis[to[i]] > dis[now] + cost[i] && wei[i]) {
        dis[to[i]] = dis[now] + cost[i];
        pre[to[i]] = now;
        pres[to[i]] = i;
        if (!vis[to[i]]) {
          que.push(to[i]);
          vis[to[i]] = 1;
        }
      }
    }
  } while (!que.empty());
  return dis[tail] != inf;
}
void f() {
  int now = tail;
  int low = inf;
  while (now != start) {
    low = min(low, wei[pres[now]]);
    now = pre[now];
  }
  used += low * dis[tail];
  flow += low;
  now = tail;
  while (now != start) {
    wei[pres[now]] -= low;
    wei[pres[now] ^ 1] += low;
    now = pre[now];
  }
}
int main() {
  cin >> n;
  start = 0, tail = 2 * n + 2;
  int s = 1;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    add(s, 2 * i, 1, 0);
    add(2 * i, 2 * i + 1, 1, -1);
    add(2 * i, 2 * i + 1, inf, 0);
    add(2 * i + 1, tail, 1, 0);
  }
  add(start, s, 4, 0);
  for (int i = 1; i <= n; i++) {
    for (int j = i + 1; j <= n; j++) {
      if (a[i] - a[j] == 1) {
        add(i * 2 + 1, j * 2, inf, 0);
        break;
      }
    }
    for (int j = i + 1; j <= n; j++) {
      if (a[j] - a[i] == 1) {
        add(i * 2 + 1, j * 2, inf, 0);
        break;
      }
    }
    for (int j = i + 1; j <= n; j++) {
      if ((a[i] - a[j]) % 7 == 0) {
        add(i * 2 + 1, j * 2, inf, 0);
        break;
      }
    }
  }
  while (spfa()) {
    f();
  }
  cout << -used << endl;
  return 0;
}