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;
const int INF = 0x7f7f7f7f;
int n, s, t, ss, tt, a[3005];
int mxf, mnc;
int ee = 1, h[5000005], nex[5000005 << 1], to[5000005 << 1], val[5000005 << 1],
    cost[5000005 << 1];
int dis[5000005], pre[5000005], last[5000005], flow[5000005];
bool vis[5000005];
void addedge(int x, int y, int z, int c) {
  nex[++ee] = h[x];
  to[ee] = y;
  val[ee] = z;
  cost[ee] = c;
  h[x] = ee;
}
bool spfa() {
  memset(dis, INF, sizeof(dis));
  memset(vis, false, sizeof(vis));
  queue<int> q;
  q.push(s);
  vis[s] = true;
  dis[s] = 0;
  flow[s] = INF;
  pre[t] = -1;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    vis[x] = false;
    for (int i = h[x]; i; i = nex[i])
      if (val[i] && dis[to[i]] > dis[x] + cost[i]) {
        dis[to[i]] = dis[x] + cost[i];
        pre[to[i]] = x;
        last[to[i]] = i;
        flow[to[i]] = min(flow[x], val[i]);
        if (!vis[to[i]]) {
          vis[to[i]] = true;
          q.push(to[i]);
        }
      }
  }
  return pre[t] != -1;
}
void mcmf() {
  mxf = mnc = 0;
  while (spfa()) {
    int now = t;
    mxf += flow[t];
    mnc += flow[t] * dis[t];
    while (now != s) {
      val[last[now]] -= flow[t];
      val[last[now] ^ 1] += flow[t];
      now = pre[now];
    }
  }
}
int main() {
  cin >> n;
  s = 0, ss = n * 3 + 1, tt = n * 3 + 2, t = n * 3 + 3;
  for (int i = 1; i <= n; cin >> a[i++])
    ;
  addedge(s, ss, 4, 0), addedge(ss, s, 0, 0);
  addedge(tt, t, 4, 0), addedge(t, tt, 0, 0);
  for (int i = 1; i <= n; i++) {
    addedge(ss, i, 4, 0), addedge(i, ss, 0, 0);
    addedge(i + n * 2, tt, 4, 0), addedge(tt, i + n * 2, 0, 0);
  }
  for (int i = 1; i < n; i++) {
    for (int j = i + 1; j <= n; j++)
      if (a[j] - a[i] == 1) {
        addedge(i + n * 2, j, INF, 0), addedge(j, i + n * 2, 0, 0);
        break;
      }
    for (int j = i + 1; j <= n; j++)
      if (a[i] - a[j] == 1) {
        addedge(i + n * 2, j, INF, 0), addedge(j, i + n * 2, 0, 0);
        break;
      }
    for (int j = i + 1; j <= n; j++)
      if (a[j] % 7 == a[i] % 7) {
        addedge(i + n * 2, j, INF, 0), addedge(j, i + n * 2, 0, 0);
        break;
      }
  }
  for (int i = 1; i <= n; i++) {
    addedge(i, i + n, 1, -1), addedge(i + n, i, 0, 1);
    addedge(i + n, i + n * 2, 1, 0), addedge(i + n * 2, i + n, 0, 0);
    addedge(i, i + n * 2, INF, 0), addedge(i + n * 2, i, 0, 0);
  }
  mcmf();
  cout << -mnc << endl;
  return 0;
}
